Alternating Sums Concerning Multiplicative Arithmetic Functions

László Tóth^{1} Department of Mathematics University of Pécs Ifjúság útja 6 7624 Pécs Hungary

                ltoth@gamma.ttk.pte.hu

Abstract

We deduce asymptotic formulas for the alternating sums ∑_{n≤x}(−1)^{n−1}f(n) and ∑_{n≤x}(−1)^{n−1} \frac{1}{f(n)}, where f is one of the following classical multiplicative arithmetic functions: Euler’s totient function, the Dedekind function, the sum-of-divisors function, the divisor function, the gcd-sum function. We also consider analogs of these functions, which are associated to unitary and exponential divisors, and other special functions. Some of our results improve the error terms obtained by Bordellès and Cloitre. We formulate certain open problems.

1 Introduction

Alternating sums and series appear in various topics of mathematics and number theory, in particular. For example, it is well-known that for s ∈ ℂ with ℜs > 1,

η(s) := ∑_{n=1}^{∞}(−1)^{n−1} \frac{1}{n^{s}} = (1 − \frac{1}{2^{s−1}} ) ζ(s),  (1)

1The present scientific contribution is dedicated to the 650th anniversary of the foundation of the University of Pécs, Hungary.

representing the alternating zeta function or Dirichlet’s eta function. Here the left-hand side is convergent for ℜs > 0, and this can be used for analytic continuation of the Riemann zeta function for ℜs > 0. See, e.g., Tenenbaum [35, Sect. II.3.2].

Bordellès and Cloitre [4] established asymptotic formulas with error terms for alternating sums

∑_{n≤x}(−1)^{n−1}f(n),  (2)

where f(n) = 1/g(n) and g belongs to a class of multiplicative arithmetic functions, including Euler’s totient function φ, the sum-of-divisors function σ and the Dedekind function ψ. It seems that there are no other results in the literature for alternating sums of type (2).

Using a different approach, also based on the convolution method, we show that for many classical multiplicative arithmetic functions f, estimates with sharp error terms for the alternating sum (2) can easily be deduced by using known results for

∑_{n≤x} f(n).  (3)

For other given multiplicative functions f, a difficulty arises, namely to estimate the coefficients of the reciprocal of a formal power series, more exactly the reciprocal of the Bell series of f for p = 2. If the coefficients of the original power series are positive and log-convex, then a result of Kaluza [16] can be used. The obtained error terms for (2) are usually the same, or slightly larger than for (3).

In this way we improve some of the error terms obtained in [4]. We also deduce estimates for other classical multiplicative functions f. As a tool, we use formulas for alternating Dirichlet series

D_{altern}(f, s) := ∑_{n=1}^{∞}(−1)^{n−1} \frac{f(n)}{n^{s}},  (4)

generalizing (1).

In the case of some other functions f, a version of Kendall’s renewal theorem (from probability theory) can be applied. Berenhaut, Allen, and Fraser obtained [3] an explicit form of Kendall’s theorem (also see [2]), but this cannot be used for the functions we deal with. We prove a new explicit Kendall-type inequality, which can be applied in some cases. As far as we know, there are no other similar applicable results to obtain better error terms in the literature. We formulate several open problems concerning the error terms of the presented asymptotic formulas.

Finally, a generalization of the alternating Dirichlet series (4) and the alternating sum (2) is discussed.

2 General results

2.1 Alternating Dirichlet series

Let

D(f, s) := ∑_{n=1}^{∞} \frac{f(n)}{n^{s}}  (5)

denote the Dirichlet series of the function f. If f is multiplicative, then it can be expanded into the Euler product

D(f, s) = ∏_{p∈ℙ} ∑_{ν=0}^{∞} \frac{f(p^{ν})}{p^{νs}}.  (6)

If f is completely multiplicative, then

∑_{ν=0}^{∞} \frac{f(p^{ν})}{p^{νs}} = (1 − \frac{f(p)}{p^{s}})^{−1}  (7)

and

D(f, s) = ∏_{p∈ℙ} (1 − \frac{f(p)}{p^{s}})^{−1} .  (8)

Proposition 1. If f is a multiplicative function, then

∑_{n=1}^{∞}(−1)^{n−1} \frac{f(n)}{n^{s}} = D(f, s) (2 (∑_{ν=0}^{∞} \frac{f(2^{ν})}{2^{νs}})^{−1} − 1) ,  (9)

and if f is completely multiplicative, then

∑_{n=1}^{∞}(−1)^{n−1} \frac{f(n)}{n^{s}} = (1 − \frac{f(2)}{2^{s−1}}) ∏_{p∈ℙ} (1 − \frac{f(p)}{p^{s}})^{−1} ,

formally or in case of convergence. Proof. We have by using (6),

∑_{n=1}^{∞}(−1)^{n−1} \frac{f(n)}{n^{s}} = − ∑_{n=1}^{∞} \frac{f(n)}{n^{s}} + 2 ∑_{\begin{matrix}n=1 \\ n odd \end{matrix}}^{∞} \frac{f(n)}{n^{s}} = −D(f, s) + 2 ∏_{\begin{matrix}p∈ℙ \\ p>2 \end{matrix}} ∑_{ν=0}^{∞} \frac{f(p^{ν})}{p^{νs}}

= −D(f, s) + 2D(f, s) (∑_{ν=0}^{∞} \frac{f(2^{ν})}{2^{νs}})^{−1} = D(f, s) (2 (∑_{ν=0}^{∞} \frac{f(2^{ν})}{2^{νs}})^{−1} − 1) .

If f is completely multiplicative, then use identities (7) and (8).

For special choices of f we obtain formulas for the alternating Dirichlet series (4). For example, let f = φ be Euler’s totient function. For every prime p,

∑_{ν=0}^{∞} \frac{φ(p^{ν})}{p^{νs}} = (1 − \frac{1}{p^{s}} ) (1 − \frac{1}{p^{s−1}} )^{−1} .  (10)

Here the left-hand side of (10) can be computed directly. However, it is more convenient to use the well-known representation of the Dirichlet series of φ (similar considerations are valid for other classical multiplicative function, as well). Namely,

∑_{n=1}^{∞} \frac{φ(n)}{n^{s}} = \frac{ζ(s − 1)}{ζ(s)} = ∏_{p∈ℙ} (1 − \frac{1}{p^{s}} ) (1 − \frac{1}{p^{s−1}} )^{−1} ,  (11)

and using the Euler product,

∑_{n=1}^{∞} \frac{φ(n)}{n^{s}} = ∏_{p∈ℙ} ∑_{ν=0}^{∞} \frac{φ(p^{ν})}{p^{νs}}.  (12)

Now a quick look at (11) and (12) gives (10). We deduce from Proposition 1 that

D_{altern}(φ, s) = \frac{ζ(s − 1)}{ζ(s)} (2 (1 − \frac{1}{2^{s}} )^{−1} (1 − \frac{1}{2^{s−1}} ) − 1) ,  (13)

which can be written as (31).

Note that the function n ↦ (−1)^{n−1} is multiplicative. Therefore, it is possible to give a direct proof of (13) (and of similar formulas, where φ is replaced by another multiplicative function) using Euler products:

∑_{n=1}^{∞} \frac{(−1)^{n−1}φ(n)}{n^{s}} = (1 − ∑_{ν=1}^{∞} \frac{φ(2^{ν})}{2^{νs}}) ∏_{\begin{matrix}p∈ℙ \\ p>2 \end{matrix}} (1 + ∑_{ν=1}^{∞} \frac{φ(p^{ν})}{p^{νs}}) ,

but computations are simpler by the previous approach.

2.2 Mean values and alternating sums

Let f be a complex-valued arithmetic function. The (asymptotic) mean value of f is

M(f) := lim_{x→∞} \frac{1}{x} ∑_{n≤x} f(n),

provided that this limit exists. Let

M_{altern}(f) := lim_{x→∞} \frac{1}{x} ∑_{n≤x}(−1)^{n−1}f(n)

denote the mean value of the function n ↦ (−1)^{n−1}f(n) (if it exists).

Proposition 2. Assume that f is a multiplicative function and

∑_{p∈ℙ} \frac{|f(p) − 1|}{p} < ∞,  ∑_{p∈ℙ} ∑_{ν=2}^{∞} \frac{|f(p^{ν})|}{p^{ν}} < ∞.  (14)

Then there exists

M(f) = ∏_{p∈ℙ} (1 − \frac{1}{p}) ∑_{ν=0}^{∞} \frac{f(p^{ν})}{p^{ν}}.

Furthermore, if ∑_{ν=0}^{∞} \frac{f(2^{ν})}{2^{ν}} ≠ 0, then there exists

M_{altern}(f) = M(f) (2 (∑_{ν=0}^{∞} \frac{f(2^{ν})}{2^{ν}})^{−1} − 1) ,  (15)

and if ∑_{ν=0}^{∞} \frac{f(2^{ν})}{2^{ν}} = 0, then M(f) = 0 and there exists

M_{altern}(f) = ∏_{\begin{matrix}p∈ℙ \\ p>2 \end{matrix}} (1 − \frac{1}{p}) ∑_{ν=0}^{∞} \frac{f(p^{ν})}{p^{ν}}.  (16)

Proof. The result for M(f) is a version of Wintner’s theorem for multiplicative functions. See Schwarz and Spilker [24, Cor. 2.3]. It easy to check that assuming (14) for f, the same conditions hold for the multiplicative function n ↦ (−1)^{n−1}f(n). We deduce that M_{altern}(f) exists and it is

M_{altern}(f) = (1 − \frac{1}{2}) (1 − ∑_{ν=1}^{∞} \frac{f(2^{ν})}{2^{ν}}) ∏_{\begin{matrix}p∈ℙ \\ p>2 \end{matrix}} (1 − \frac{1}{p}) ∑_{ν=0}^{∞} \frac{f(p^{ν})}{p^{ν}}.  (17)

Now if t := ∑_{ν=1}^{∞} \frac{f(2^{ν})}{2^{ν}} ≠ −1, then

M_{altern}(f) = \frac{1 − t}{1 + t} ∏_{p∈ℙ} (1 − \frac{1}{p}) ∑_{ν=0}^{∞} \frac{f(p^{ν})}{p^{ν}},

which is (15). If t = −1, then (17) gives (16).

Application 3. Let f be multiplicative such that f(p) = 1, f(p^{2}) = −6, f(p^{ν}) = 0 for every p ∈ ℙ and every ν ≥ 3. Here ∑_{ν=0}^{∞} \frac{f(2^{ν})}{2^{ν}} = 1 + \frac{1}{2} − \frac{6}{4} = 0. Using Proposition 2 we deduce that M(f) = 0 and there exists

lim_{x→∞} \frac{1}{x} ∑_{n≤x}(−1)^{n−1}f(n) = ∏_{\begin{matrix}p∈ℙ \\ p>2 \end{matrix}} (1 − \frac{1}{p}) (1 + \frac{1}{p} − \frac{6}{p^{2}} ) = ∏_{\begin{matrix}p∈ℙ \\ p>2 \end{matrix}} (1 − \frac{7}{p^{2}} + \frac{6}{p^{3}} ) ≠ 0.

The following result is similar. Let

L(f) := lim_{x→∞} \frac{1}{log x} ∑_{n≤x} \frac{f(n)}{n}

denote the logarithmic mean value of f and, assuming that f is non-vanishing, let

\overline{L}(f) := lim_{x→∞} \frac{1}{log x} ∑_{n≤x} \frac{1}{f(n)},

\overline{L}_{altern}(f) := lim_{x→∞} \frac{1}{log x} ∑_{n≤x}(−1)^{n−1} \frac{1}{f(n)},

provided that the limits exist. Proposition 4. Assume that f is a non-vanishing multiplicative function and

∑_{p∈ℙ} |\frac{1}{f(p)} − \frac{1}{p}| < ∞,  ∑_{p∈ℙ} ∑_{ν=2}^{∞} \frac{1}{|f(p^{ν})|} < ∞.  (18)

Then there exists

\overline{L}(f) = ∏_{p∈ℙ} (1 − \frac{1}{p}) ∑_{ν=0}^{∞} \frac{1}{f(p^{ν})}.

Furthermore, if ∑_{ν=0}^{∞} \frac{1}{f(2^{ν})} ≠ 0, then there exists

\overline{L}_{altern}(f) = \overline{L}(f) (2 (∑_{ν=0}^{∞} \frac{1}{f(2^{ν})})^{−1} − 1) ,

and if ∑_{ν=0}^{∞} \frac{1}{f(2^{ν})} = 0, then \overline{L}(f) = 0 and there exists

\overline{L}_{altern}(f) = ∏_{\begin{matrix}p∈ℙ \\ p>2 \end{matrix}} (1 − \frac{1}{p}) ∑_{ν=0}^{∞} \frac{1}{f(p^{ν})}.

Proof. Apply Proposition 2 for f(n) := \frac{n}{f(n)} and use the following property: If the mean value M(f) exists, then the logarithmic mean value L(f) exists as well, and is equal to M(f). See Hildebrand [14, Thm. 2.13]. Application 5. It follows from Proposition 4 that

\overline{L}(φ) = lim_{x→∞} \frac{1}{log x} ∑_{n≤x} \frac{1}{φ(n)} = ∏_{p∈ℙ} (1 − \frac{1}{p}) ∑_{ν=0}^{∞} \frac{1}{φ(p^{ν})} = \frac{ζ(2)ζ(3)}{ζ(6)},

which is well-known, and

lim_{x→∞} \frac{1}{log x} ∑_{n≤x}(−1)^{n−1} \frac{1}{φ(n)} = \overline{L}(φ) (2 (∑_{ν=0}^{∞} \frac{1}{φ(2^{ν})})^{−1} − 1) = −\frac{ζ(2)ζ(3)}{3ζ(6)},

obtained by Bordellès and Cloitre [4]. Conditions (18) were refined in [4] to deduce asymptotic formulas with error terms for alternating sums of reciprocals of a class of multiplicative arithmetic functions, including Euler’s totient function.

2.3 Method to obtain asymptotic formulas

Assume that f is a nonzero complex-valued multiplicative function. Consider the formal power series

S_{f}(x) := ∑_{ν=0}^{∞} a_{ν}x^{ν},

where a_{ν} = f(2^{ν}) (ν ≥ 0), a_{0} = f(1) = 1. Note that S_{f}(x) is the Bell series of the function f for the prime p = 2. See, e.g., Apostol [1, Ch. 2]. Let

\overline{S}_{f}(x) := ∑_{ν=0}^{∞} b_{ν}x^{ν}

be its formal reciprocal power series. Here the coefficients b_{ν} are given by b_{0} = 1 and ∑_{j=0}^{ν} a_{j}b_{ν−j} = 0 (ν ≥ 1). If both series S_{f}(x) and \overline{S}_{f}(x) converge for an x ∈ ℂ, then S_{f}(x)\overline{S}_{f}(x) = 1. In particular, if r_{f} and \overline{r}_{f} are the radii of convergence of S_{f}(x), respectively \overline{S}_{f}(x), then S_{f}(x)\overline{S}_{f}(x) = 1 for every x ∈ ℂ such that |x| < min(r_{f}, \overline{r}_{f}).

It follows from (9) that the convolution identity

(−1)^{n−1}f(n) = ∑_{dj=n} h_{f}(d)f(j) (n ≥ 1)  (19)

holds, where the function h_{f} is multiplicative, h_{f}(p^{ν}) = 0 if p > 2, ν ≥ 1 and h_{f}(2^{ν}) = 2b_{ν} (ν ≥ 1), h_{f}(1) = 2b_{0} − 1 = 1.

Therefore, by the convolution method,

∑_{n≤x}(−1)^{n−1}f(n) = ∑_{d≤x} h_{f}(d) ∑_{j≤x/d} f(j),  (20)

which leads to a good estimate for (2) if an asymptotic formula for ∑_{n≤x} f(n) is known and if the coefficients b_{ν} of above can be well estimated. Note that, according to (19) and (9),

∑_{n=1}^{∞} \frac{h_{f}(n)}{n^{s}} = \frac{2}{S_{f}(1/2^{s})} − 1,  (21)

provided that both S_{f}(1/2^{s}) and \overline{S}_{f}(1/2^{s}) converge. By differentiating,

∑_{n=1}^{∞} \frac{h_{f}(n) log n}{n^{s}} = −\frac{log 2}{2^{s−1}} · \frac{S′_{f}(1/2^{s})}{S_{f}(1/2^{s})^{2}} ,  (22)

assuming that |1/2^{s}| < min(r_{f}, \overline{r}_{f}). Identities (21) and (22) will be used in applications.

2.4 Two general asymptotic formulas

We prove two general results that will be applied for several special functions in Section 4.

Proposition 6. Let f be a multiplicative function. Assume that (i) there exists a constant C_{f} such that

∑_{n≤x} f(n) = C_{f}x^{2} + O (xR_{f}(x)) ,

where 1 ≪ R_{f}(x) = o(x) as x → ∞, and R_{f}(x) is nondecreasing; (ii) S_{f}(1/4) converges; (iii) the sequence (b_{ν})_{ν≥0} of coefficients of the reciprocal power series \overline{S}_{f}(x) is bounded. Then

∑_{n≤x}(−1)^{n−1}f(n) = C_{f} (\frac{2}{S_{f}(1/4)} − 1) x^{2} + O (xR_{f}(x)) .

Proof. According to (20),

∑_{n≤x}(−1)^{n−1}f(n) = ∑_{d≤x} h_{f}(d) (C_{f} \frac{x^{2}}{d^{2}} + O (\frac{x}{d}R_{f}(x/d)))

= C_{f}x^{2} ∑_{d≤x} \frac{h_{f}(d)}{d^{2}} + O (xR_{f}(x) ∑_{d≤x} \frac{|h_{f}(d)|}{d}) .

Since the sequence (b_{ν})_{ν≥0} is bounded, the function h_{f} is bounded. Moreover, the sum

∑_{d≤x} \frac{|h_{f}(d)|}{d} = ∑_{d=2^{ν}≤x} \frac{|h_{f}(2^{ν})|}{2^{ν}} ≪ ∑_{2^{ν}≤x} \frac{|b_{ν}|}{2^{ν}}

is bounded, as well. Note that S_{f}(1/4) and \overline{S}_{f}(1/4) both converge by conditions (ii) and (iii). We deduce, by using (21) for s = 2, that

∑_{n≤x}(−1)^{n−1}f(n) = C_{f}x^{2} ∑_{d=1}^{∞} \frac{h_{f}(d)}{d^{2}} + O (x^{2} ∑_{d>x} \frac{1}{d^{2}} ) + O (xR_{f}(x))

= C_{f}x^{2} (\frac{2}{S_{f}(1/4)} − 1) + O (xR_{f}(x)) .

Proposition 7. Let f be a nonvanishing multiplicative function. Assume that (i) there exist constants D_{f} and E_{f} such that

∑_{n≤x} \frac{1}{f(n)} = D_{f}(log x + E_{f}) + O (x^{−1}R_{1/f}(x)) ,  (23)

where 1 ≪ R_{1/f}(x) = o(x) as x → ∞, and R_{1/f}(x) is nondecreasing; (ii) the radius of convergence of the series S_{1/f}(x) is r_{1/f} > 1; (iii) the coefficients b_{ν} of the reciprocal power series \overline{S}_{1/f}(x) satisfy b_{ν} ≪ M^{ν} as ν → ∞,

where 0 < M < 1 is a real number.

Then

∑_{n≤x}(−1)^{n−1} \frac{1}{f(n)} = D_{f} ((\frac{2}{S_{1/f}(1)} − 1) (log x + E_{f}) + 2(log 2)\frac{S′_{1/f}(1)}{S_{1/f}(1)^{2}} ) + O (T_{1/f}(x)) ,

(24)

where

T_{1/f}(x) = \begin{cases}x^{−1}R_{1/f}(x), & if 0 < M < \frac{1}{2}; \\ x^{−1}R_{1/f}(x) log x, & if M = \frac{1}{2}; \\ x^{log M/ log 2} max(log x, R_{1/f}(x)), & if \frac{1}{2} < M < 1. \end{cases}  (25)

Proof. According to (20) we deduce that

∑_{n≤x}(−1)^{n−1} \frac{1}{f(n)} = ∑_{d≤x} h_{1/f}(d) ∑_{j≤x/d} \frac{1}{f(j)}

= ∑_{d≤x} h_{1/f}(d) (D_{f} (log \frac{x}{d} + E_{f}) + O ((x/d)^{−1}R_{1/f}(x/d)))

= D_{f}(log x + E_{f}) ∑_{d≤x} h_{1/f}(d) − D_{f} ∑_{d≤x} h_{1/f}(d) log d

+ O (x^{−1}R_{1/f}(x) ∑_{d≤x} d|h_{1/f}(d)|) .

That is,

∑_{n≤x}(−1)^{n−1} \frac{1}{f(n)} = D_{f}(log x + E_{f}) ∑_{d=1}^{∞} h_{1/f}(d) + O (log x ∑_{d>x} |h_{1/f}(d)|)

−D_{f} ∑_{d=1}^{∞} h_{1/f}(d) log d + O (∑_{d>x} |h_{1/f}(d)| log d) + O (x^{−1}R_{1/f}(x) ∑_{d≤x} d|h_{1/f}(d)|) . (26)

Note that min(r_{1/f}, \overline{r}_{1/f}) > 1 by conditions (ii) and (iii). By using (21) and (22) for s = 0,

∑_{d=1}^{∞} h_{1/f}(d) = \frac{2}{S_{1/f}(1)} − 1,

∑_{d=1}^{∞} h_{1/f}(d) log d = −2(log 2)\frac{S′_{1/f}(1)}{S_{1/f}(1)^{2}} .

Furthermore,

∑_{d>x} |h_{1/f}(d)| = ∑_{d=2^{ν}>x} |h_{1/f}(2^{ν})| ≪ ∑_{2^{ν}>x} |b_{ν}| ≪ ∑_{2^{ν}>x} M^{ν} ≪ x^{log M/ log 2},

∑_{d>x} |h_{1/f}(d)| log d ≪ ∑_{2^{ν}>x} ν|b_{ν}| ≪ ∑_{2^{ν}>x} νM^{ν} ≪ x^{log M/ log 2} log x,

∑_{d≤x} d|h_{1/f}(d)| = ∑_{2^{ν}≤x} 2^{ν}|b_{ν}| ≪ ∑_{ν≤log x/ log 2}(2M)^{ν},

where the latter sum is bounded if 0 < M < 1/2, it is ≪ log x if M = 1/2, and is ≪ x^{1+log M/ log 2} if 1/2 < M < 1.

Inserting these into (26), the proof is complete.

3 Estimates on coefficients of reciprocal power series

As mentioned in Section 2.3, in order to deduce sharp error terms for alternating sums (2) we need good estimates for the coefficients b_{ν} of the power series \overline{S}_{f}(x).

3.1 Theorem of Kaluza

In many (nontrivial) cases the next result can be used.

Lemma 8. Let ∑_{ν=0}^{∞} a_{ν}x^{ν} be a power series such that a_{ν} > 0 (ν ≥ 0) and the sequence (a_{ν})_{ν≥0} is log-convex, that is a_{ν}^{2} ≤ a_{ν−1}a_{ν+1} (ν ≥ 1). Then for the coefficients b_{ν} of the (formal) reciprocal power series ∑_{ν=0}^{∞} b_{ν}x^{ν} one has b_{0} = 1/a_{0} > 0 and

−\frac{1}{a_{0}^{2}} a_{ν} ≤ b_{ν} ≤ 0 for all ν ≥ 1.

Proof. The property that b_{ν} ≤ 0 for all ν ≥ 1 is the theorem of Kaluza [16, Satz 3]. See [6] for a short direct proof of it. Furthermore, we have

b_{ν} = −\frac{1}{a_{0}^{2}} a_{ν} − \frac{1}{a_{0}} ∑_{j=1}^{ν−1} a_{j}b_{ν−j} ≥ −\frac{1}{a_{0}^{2}} a_{ν} (ν ≥ 1).

For example, consider the sum-of-divisors function σ, where σ(2^{ν}) = 2^{ν+1} − 1 for every ν ≥ 0. The sequence (\frac{1}{2^{ν+1}−1})_{ν≥0} is log-convex. This property allows us to apply Lemma 8 to obtain the estimate of Theorem 23 for the alternating sum ∑_{n≤x}(−1)^{n−1} \frac{1}{σ(n)}.

3.2 Kendall’s renewal theorem

Another related result is Kendall’s renewal theorem. Disregarding the probabilistic context, it can be stated as follows. See Berenhaut, Allen, and Fraser [3, Thm. 1.1].

Lemma 9. Let ∑_{ν=0}^{∞} a_{ν}x^{ν} be a power series such that (a_{ν})_{ν≥0} is nonincreasing, a_{0} = 1, a_{ν} ≥ 0 (ν ≥ 1) and a_{ν} ≪ q^{ν} as ν → ∞, where 0 < q < 1 is a real number. Then there exists 0 < s < 1, s real, such that for the coefficients b_{ν} of the reciprocal power series one has b_{ν} ≪ s^{ν} as ν → ∞.

We deduce the next result:

Corollary 10. Let f be a positive multiplicative function. Assume that (i) asymptotic formula (23) is valid with 1 ≪ R_{1/f}(x) ≪ x^{ε} as x → ∞, for every ε > 0; (ii) the sequence (1/f(2^{ν}))_{ν≥0} is nonincreasing and 1/f(2^{ν}) ≪ q^{ν} as ν → ∞, where

0 < q < 1 is a real number.

Then the asymptotic formula (24) holds for ∑_{n≤x}(−1)^{n−1} \frac{1}{f(n)}, with error term

T_{1/f}(x) = x^{−u} max(log x, R_{1/f}(x)) ≪ x^{−u_{1}}

for some u, u_{1} > 0.

Proof. This is a direct consequence of Proposition 7 and Lemma 9, applied for a_{ν} = \frac{1}{f(2^{ν})}. Note that the radius of convergence of the series S_{1/f}(x) is > 1 by condition (ii).

In the case of the sum-of-unitary-divisors function σ^{∗} we have a_{ν} = σ^{∗}(2^{ν}) = 2^{ν} + 1 for every ν ≥ 1 and a_{0} = σ^{∗}(1) = 1. The sequence (\frac{1}{a_{ν}} )_{ν≥0} is not log-convex. Lemma 8 cannot

be used to estimate the alternating sum ∑_{n≤x}(−1)^{n−1} \frac{1}{σ^{∗}(n)}. At the same time, Corollary 10, with t = 2 furnishes an asymptotic formula. See Section 4.9.

An explicit form of Lemma 9 (Kendall’s theorem) was proved in [3, Thm. 1.2]. However, it is restricted to the values 0 < q < 0.32, and cannot be applied for the above special case, where q = 1/2. To find the optimal value of s for pairs (A, q) such that a_{ν} ≤ Aq^{ν} (ν ≥ 1), not satisfying assumptions of [3, Thm. 1.2] was formulated by Berenhaut, Abernathy, Fan, and Foley [2, Open question 5.4].

We prove a new explicit Kendall-type inequality, based on the following lemma.

Lemma 11. Let ∑_{ν=0}^{∞} a_{ν}x^{ν} be a power series such that a_{0} = 1. Then for the coefficients b_{ν} of the reciprocal power series ∑_{ν=0}^{∞} b_{ν}x^{ν} one has b_{0} = 1 and

b_{ν} = ∑_{k=1}^{ν}(−1)^{k} ∑_{\begin{matrix}j_{1},...,j_{k}≥1 \\ j_{1}+···+j_{k}=ν \end{matrix}} a_{j_{1}} ⋯ a_{j_{k}}  (27)

= ∑_{\begin{matrix}t_{1},...,t_{ν}≥0 \\ t_{1}+2t_{2}+···+νt_{ν}=ν \end{matrix}}(−1)^{t_{1}+···+t_{ν}} \binom{t_{1} + ⋯ + t_{ν}}{t_{1}, …, t_{ν}}a_{1}^{t_{1}} ⋯ a_{ν}^{t_{ν}}  (28)

for every ν ≥ 1, where \binom{t_{1}+···+t_{ν}}{t_{1},...,t_{ν}} are the multinomial coefficients.

Here formula (28) is well known, and it has been recovered several times. See, e.g., [19, Lemma 4]. However, we were not able to find its equivalent version (27) in the literature. For the sake of completeness we present their proofs.

Proof. Using the geometric series formula (1 + x)^{−1} = ∑_{n=0}^{∞}(−1)^{n}x^{n} and the multinomial theorem, we immediately have

(1 + ∑_{ν=1}^{∞} a_{ν}x^{ν})^{−1} = ∑_{t=0}^{∞}(−1)^{t} (∑_{ν=1}^{∞} a_{ν}x^{ν})^{t}

= ∑_{ν=0}^{∞} x^{ν} ∑_{\begin{matrix}t_{1},...,t_{ν}≥0 \\ t_{1}+2t_{2}+···+νt_{ν}=ν \end{matrix}}(−1)^{t_{1}+···+t_{ν}} \binom{t_{1} + ⋯ + t_{ν}}{t_{1}, …, t_{ν}}a_{1}^{t_{1}} ⋯ a_{ν}^{t_{ν}},

giving (28). Furthermore, fix ν ≥ 1. By grouping the terms in (28) according to the values k = t_{1} + ⋯ + t_{ν}, where 1 ≤ k ≤ ν, we have

b_{ν} = ∑_{k=1}^{ν}(−1)^{k} ∑_{\begin{matrix}t_{1},...,t_{ν}≥0 \\ t_{1}+2t_{2}+···+νt_{ν}=ν \\ t_{1}+···+t_{ν}=k \end{matrix}} \binom{t_{1} + ⋯ + t_{ν}}{t_{1}, …, t_{ν}}a_{1}^{t_{1}} ⋯ a_{ν}^{t_{ν}}.

Now, identity (27) follows if we show that

∑_{\begin{matrix}j_{1},...,j_{k}≥1 \\ j_{1}+···+j_{k}=ν \end{matrix}} a_{j_{1}} ⋯ a_{j_{k}} = ∑_{\begin{matrix}t_{1},...,t_{ν}≥0 \\ t_{1}+2t_{2}+···+νt_{ν}=ν \\ t_{1}+···+t_{ν}=k \end{matrix}} \binom{t_{1} + ⋯ + t_{ν}}{t_{1}, …, t_{ν}}a_{1}^{t_{1}} ⋯ a_{ν}^{t_{ν}}.  (29)

But (29) is immediate by starting with its left-hand side and denoting by t_{1}, …, t_{ν} the number of values j_{1}, …, j_{k} which are equal to 1, …, ν, respectively.

Proposition 12. Assume that ∑_{ν=0}^{∞} a_{ν}x^{ν} is a power series such that a_{0} = 1 and |a_{ν}| ≤ Aq^{ν} (ν ≥ 1) for some absolute constants A, q > 0. Then for the coefficients b_{ν} of the reciprocal power series one has

|b_{ν}| ≤ Aq^{ν}(A + 1)^{ν−1} (ν ≥ 1).  (30)

Proof. By identity (27) and the assumption |a_{ν}| ≤ Aq^{ν} (ν ≥ 1) we immediately have

|b_{ν}| ≤ ∑_{k=1}^{ν} A^{k} ∑_{\begin{matrix}j_{1},...,j_{k}≥1 \\ j_{1}+...+j_{k}=ν \end{matrix}} q^{j_{1}+···+j_{k}} = q^{ν} ∑_{k=1}^{ν} A^{k} ∑_{\begin{matrix}j_{1},...,j_{k}≥1 \\ j_{1}+...+j_{k}=ν \end{matrix}} 1

= q^{ν} ∑_{k=1}^{ν} A^{k}\binom{ν − 1}{k − 1} = Aq^{ν}(A + 1)^{ν−1},

as asserted.

Note that (30) is an explicit Kendall-type inequality provided that q(A + 1) < 1, in particular if q ≤ 1/2 and A < 1.

Corollary 13. Let f be a positive multiplicative function such that (i) asymptotic formula (23) is valid with 1 ≪ R_{1/f}(x) = o(x) as x → ∞; (ii) 1/f(2^{ν}) ≤ Aq^{ν} (ν ≥ 1), where A, q > 0 are fixed real constants satisfying M :=

q(A + 1) < 1.

Then the asymptotic formula (24) holds for ∑_{n≤x}(−1)^{n−1} \frac{1}{f(n)}, with error term (25).

Proof. This follows from Propositions 7 and 12. Note that the radius of convergence of the series S_{1/f}(x) is > 1 by condition (ii).

We will apply Corollary 13 for the sum-of-bi-unitary-divisors function σ^{∗∗}. See Section 4.13.

4 Results for classical functions

In this section, we investigate alternating sums for classical multiplicative functions. We refer to Apostol [1], Hildebrand [14], and McCarthy [18] for the basic properties of these functions. See Gould and Shonhiwa [12] for a list of Dirichlet series of special arithmetic functions.

4.1 Euler’s totient function

First consider Euler’s φ function, where φ(n) = n ∏_{p|n} (1 − \frac{1}{p}) (n ≥ 1).

Proposition 14.

∑_{n=1}^{∞}(−1)^{n−1} \frac{φ(n)}{n^{s}} = \frac{2^{s} − 3}{2^{s} − 1} · \frac{ζ(s − 1)}{ζ(s)} (ℜs > 2).  (31)

Proof. This was explained in Section 2.1, formula (31) follows at once from (13).

Theorem 15.

∑_{n≤x}(−1)^{n−1}φ(n) = \frac{1}{π^{2}} x^{2} + O (x(log x)^{2/3}(log log x)^{4/3}) .  (32)

Proof. Apply Proposition 6 for f = φ. It is known that

∑_{n≤x} φ(n) = \frac{3}{π^{2}} x^{2} + O (x(log x)^{2/3}(log log x)^{4/3}) ,

which is the best error term known to date, due to Walfisz [39, Satz 1, p. 144]. Furthermore,

S_{φ}(x) = ∑_{ν=0}^{∞} φ(2^{ν})x^{ν} = 1 + ∑_{ν=1}^{∞} 2^{ν−1}x^{ν} = \frac{1 − x}{1 − 2x} (|x| < \frac{1}{2})

(also see (10)). We obtain that the reciprocal power series is

\overline{S}_{φ}(x) = \frac{1 − 2x}{1 − x} = 1 − ∑_{ν=1}^{∞} x^{ν} (|x| < 1),

for which the coefficients are b_{0} = 1, b_{ν} = −1 (ν ≥ 1), forming a bounded sequence. The coefficient of the main term in (32) is

C_{φ} (\frac{2}{S_{φ}(1/4)} − 1) = \frac{3}{π^{2}} · \frac{1}{3} = \frac{1}{π^{2}} .

Remark 16. To find the corresponding constant to be multiplied by C_{φ} = 3/π^{2} observe that by (19), (21) and (31),

\frac{2}{S_{φ}(1/4)} − 1 = [\frac{2^{s} − 3}{2^{s} − 1}]_{s=2} = \frac{1}{3},

and similarly for other classical multiplicative functions, if we have the representation of their alternating Dirichlet series.

Theorem 17.

∑_{n≤x}(−1)^{n−1} \frac{1}{φ(n)} = −\frac{A}{3} (log x + γ − B − \frac{8}{3} log 2) + O (x^{−1}(log x)^{5/3}) ,  (33)

where γ is Euler’s constant and

A = \frac{ζ(2)ζ(3)}{ζ(6)} = \frac{315ζ(3)}{2π^{4}},  B = ∑_{p∈ℙ} \frac{log p}{p^{2} − p + 1}.  (34)

The result (33) improves the error term O (x^{−1}(log x)^{3}) obtained by Bordellès and Cloitre [4, Cor. 4, (i)].

Proof. Apply Proposition 7 for f = φ. The asymptotic formula

∑_{n≤x} \frac{1}{φ(n)} = A (log x + γ − B) + O (x^{−1}(log x)^{2/3})

with constants A and B defined by (34) and with the weaker error term O (x^{−1} log x) goes back to the work of Landau. See [9, Thm. 1.1]. The error term above was obtained by Sitaramachandrarao [25].

Now

S_{1/φ}(x) = ∑_{ν=0}^{∞} \frac{x^{ν}}{φ(2^{ν})} = 1 + ∑_{ν=1}^{∞} \frac{x^{ν}}{2^{ν−1}} = \frac{2 + x}{2 − x} (|x| < 2),

\overline{S}_{1/φ}(x) = \frac{1 − x/2}{1 + x/2} = 1 + ∑_{ν=1}^{∞}(−1)^{ν} \frac{x^{ν}}{2^{ν−1}} (|x| < 2);

hence b_{ν} ≪ 2^{−ν} and choose M = 1/2. Using that S_{1/φ}(1) = 3 and S′_{1/φ}(1) = 4, the proof is complete.

4.2 Dedekind function

The Dedekind function ψ is given by ψ(n) = n ∏_{p|n} (1 + \frac{1}{p}) (n ≥ 1).

Proposition 18.

∑_{n=1}^{∞}(−1)^{n−1} \frac{ψ(n)}{n^{s}} = \frac{2^{s} − 5}{2^{s} + 1} · \frac{ζ(s)ζ(s − 1)}{ζ(2s)} (ℜs > 2),  (35)

Proof. It is well-known that

∑_{n=1}^{∞} \frac{ψ(n)}{n^{s}} = \frac{ζ(s)ζ(s − 1)}{ζ(2s)} (ℜs > 2),

and (35) follows like (31), by using Proposition 1.

Theorem 19.

∑_{n≤x}(−1)^{n−1}ψ(n) = −\frac{3}{2π^{2}} x^{2} + O (x(log x)^{2/3}) ,  (36)

Proof. Apply Proposition 6 for f = ψ. It is known that

∑_{n≤x} ψ(n) = \frac{15}{2π^{2}} x^{2} + O (x(log x)^{2/3}) ,

the best estimate up to now. See Walfisz [39, Satz 2, p. 100]. Here

S_{ψ}(x) = ∑_{ν=0}^{∞} ψ(2^{ν})x^{ν} = 1 + 3 ∑_{ν=1}^{∞} 2^{ν−1}x^{n} = \frac{1 + x}{1 − 2x} (|x| < \frac{1}{2}) .

We obtain that the reciprocal power series is

\overline{S}_{ψ}(x) = \frac{1 − 2x}{1 + x} = 1 + 3 ∑_{ν=1}^{∞}(−1)^{ν}x^{ν} (|x| < 1),

for which the coefficients are b_{0} = 1, b_{ν} = 3(−1)^{ν} (ν ≥ 1), forming a bounded sequence. The coefficient of the main term in (36) is

C_{ψ} (\frac{2}{S_{ψ}(1/4)} − 1) = \frac{15}{2π^{2}} · (−\frac{1}{5}) = −\frac{3}{2π^{2}} .

Theorem 20.

∑_{n≤x}(−1)^{n−1} \frac{1}{ψ(n)} = \frac{C}{5} (log x + γ + D + \frac{24}{5} log 2) + O (x^{−1}(log x)^{2/3}(log log x)^{4/3}) , (37)

where γ is Euler’s constant and

C = ∏_{p∈ℙ} (1 − \frac{1}{p(p + 1)}) ,  D = ∑_{p∈ℙ} \frac{log p}{p^{2} + p − 1}.  (38)

The result (37) improves the error term O (x^{−1}(log x)^{2}) obtained by Bordellès and Cloitre [4, Cor. 4, (iii)]. The constant C ≐ 0.704442 is sometimes called the carefree constant, and its digits form the sequence \underline{A065463} in Sloane’s Online Encyclopedia of Integer Sequences (OEIS) [31]. Also see Finch [11, Sect. 2.5.1].

Proof. Apply Proposition 7 for f = ψ. The asymptotic formula

∑_{n≤x} \frac{1}{ψ(n)} = C (log x + γ + D) + O (x^{−1}(log x)^{2/3}(log log x)^{4/3}) ,

where C and D are the constants given by (38), is due to Sita Ramaiah and Suryanarayana [29, Cor. 4.2].

Furthermore,

S_{1/ψ}(x) = ∑_{ν=0}^{∞} \frac{x^{ν}}{ψ(2^{ν})} = 1 + \frac{2}{3} ∑_{ν=1}^{∞} \frac{x^{ν}}{2^{ν}} = \frac{6 − x}{3(2 − x)} (|x| < 2),

\overline{S}_{1/ψ}(x) = \frac{1 − x/2}{1 − x/6} = 1 − 2 ∑_{ν=1}^{∞} \frac{x^{ν}}{6^{ν}} (|x| < 6),

which shows that

b_{ν} = −\frac{2}{6^{ν}} (ν ≥ 1).

Hence b_{ν} ≪ 6^{−ν} and choose M = 1/6. Using that S_{1/ψ}(1) = \frac{5}{3} and S′_{1/ψ}(1) = \frac{4}{3}, we deduce (37).

4.3 Sum-of-divisors function

Consider the function σ(n) = ∑_{d|n} d (n ≥ 1). Proposition 21.

∑_{n=1}^{∞}(−1)^{n−1} \frac{σ(n)}{n^{s}} = (1 − \frac{6}{2^{s}} + \frac{4}{2^{2s}} ) ζ(s)ζ(s − 1) (ℜs > 2).  (39)

Note that (−1)^{n−1}σ(n) is sequence \underline{A143348} in the OEIS [31], where identity (39) is given. Proof. Use the familiar formula

∑_{n=1}^{∞} \frac{σ(n)}{n^{s}} = ζ(s)ζ(s − 1) (ℜs > 2)

and Proposition 1. Theorem 22.

∑_{n≤x}(−1)^{n−1}σ(n) = −\frac{π^{2}}{48}x^{2} + O (x(log x)^{2/3}) .  (40)

Proof. Apply Proposition 6 for f = σ. It is known that

∑_{n≤x} σ(n) = \frac{π^{2}}{12}x^{2} + O (x(log x)^{2/3}) ,

the best up to now. See Walfisz [39, Satz 4, p. 99]. Here

S_{σ}(x) = ∑_{ν=0}^{∞} σ(2^{ν})x^{ν} = ∑_{ν=0}^{∞}(2^{ν+1} − 1)x^{n} = \frac{1}{(1 − x)(1 − 2x)} (|x| < \frac{1}{2}) .

Hence

\overline{S}_{σ}(x) = (1 − x)(1 − 2x) = 1 − 3x + 2x^{2},

for which the coefficients are b_{0} = 1, b_{1} = −3, b_{2} = 2, b_{ν} = 0 (ν ≥ 3). The coefficient of the main term in (40) is from (39),

\frac{π^{2}}{12} [1 − \frac{6}{2^{s}} + \frac{4}{2^{2s}} ]_{s=2} = −\frac{π^{2}}{48}.

The following asymptotic formula is due to Sita Ramaiah and Suryanarayana [29, Cor. 4.1]:

∑_{n≤x} \frac{1}{σ(n)} = E (log x + γ + F) + O (x^{−1}(log x)^{2/3}(log log x)^{4/3}) ,  (41)

where γ is Euler’s constant,

E = ∏_{p∈ℙ} α(p), F = ∑_{p∈ℙ} \frac{(p − 1)^{2}β(p) log p}{pα(p)},

α(p) = (1 − \frac{1}{p}) ∑_{ν=0}^{∞} \frac{1}{σ(p^{ν})} = 1 − \frac{(p − 1)^{2}}{p} ∑_{j=1}^{∞} \frac{1}{(p^{j} − 1)(p^{j+1} − 1)},

β(p) = ∑_{j=1}^{∞} \frac{j}{(p^{j} − 1)(p^{j+1} − 1)}.

We prove the next result:

Theorem 23.

∑_{n≤x}(−1)^{n−1} \frac{1}{σ(n)} = E ((\frac{2}{K} − 1) (log x + γ + F) + 2(log 2)\frac{K′}{K^{2}} )  (42)

+O (x^{−1}(log x)^{5/3}(log log x)^{4/3}) ,

where

K = ∑_{j=0}^{∞} \frac{1}{2^{j+1} − 1},  K′ = ∑_{j=1}^{∞} \frac{j}{2^{j+1} − 1}.  (43)

The result (42) improves the error term O (x^{−1}(log x)^{4}) obtained by Bordellès and Cloitre [4, Cor. 4, (v)]. Here K ≐ 1.606695 is the Erdős-Borwein constant, known to be irrational. See sequence \underline{A065442} in the OEIS [31].

Proof. Apply Proposition 7 for f = σ, using formula (41). Now

S_{1/σ}(x) = ∑_{ν=0}^{∞} \frac{x^{ν}}{σ(2^{ν})} = ∑_{ν=0}^{∞} \frac{x^{ν}}{2^{ν+1} − 1}

and S_{1/σ}(1) = K. Note that S′_{1/ψ}(1) = K′, given above.

The coefficients b_{ν} of the reciprocal power series are b_{0} = 1, b_{1} = −\frac{1}{3}, b_{2} = −\frac{2}{63}, b_{3} = −\frac{8}{945}, etc. Observe that the sequence (\frac{1}{2^{ν+1}−1})_{ν≥0} is log-convex. Therefore, according to Lemma 8,

−\frac{1}{2^{ν+1} − 1} ≤ b_{ν} ≤ 0 (ν ≥ 1),

which shows that b_{ν} ≪ 2^{−ν} and we can choose M = 1/2.

4.4 Divisor function

Now consider another classical function, the divisor function τ(n) = ∑_{d|n} 1 (n ≥ 1). Using the familiar formula

∑_{n=1}^{∞} \frac{τ(n)}{n^{s}} = ζ^{2}(s) (ℜs > 1),

and Proposition 1 we deduce Proposition 24.

∑_{n=1}^{∞}(−1)^{n−1} \frac{τ(n)}{n^{s}} = (1 − \frac{4}{2^{s}} + \frac{2}{2^{2s}} ) ζ^{2}(s) (ℜs > 1).

By similar considerations we also have Proposition 25.

∑_{n=1}^{∞}(−1)^{n−1} \frac{1}{τ(n)n^{s}} = (\frac{1}{2^{s−1}} (log (1 − \frac{1}{2^{s}} ))^{−1} + 1) ∏_{p∈ℙ} p^{s} log (1 − \frac{1}{p^{s}} ) (ℜs > 1).

Theorem 26.

∑_{n≤x}(−1)^{n−1}τ(n) = −\frac{1}{2}x log x + (\frac{1}{2} − γ + log 2) x + O (x^{θ+ε}) ,

where θ is the best exponent in Dirichlet’s divisor problem. Proof. Proposition 6 cannot be applied. Using that

∑_{n≤x} τ(n) = x log x + (2γ − 1)x + O (x^{θ+ε})

the result follows by similar arguments.

Note that the actual best result for θ is θ = 131/416 ≐ 0.314903, due to Huxley [15]. Now we consider the following result, which goes back to the work of Ramanujan [21,

Eq. (7)]. See Wilson [40, Sect. 3] for its proof:

∑_{n≤x} \frac{1}{τ(n)} = x ∑_{j=1}^{N} \frac{A_{j}}{(log x)^{j−1/2}} + O (\frac{x}{(log x)^{N+1/2}} ) ,

valid for every real x ≥ 2 and every fixed integer N ≥ 1 where A_{j} (1 ≤ j ≤ N) are computable constants,

A_{1} = \frac{1}{\sqrt{π}} ∏_{p∈ℙ} (\sqrt{p^{2} − p} log (\frac{p}{p − 1})) .

We prove

Theorem 27.

∑_{n≤x}(−1)^{n−1} \frac{1}{τ(n)} = x ∑_{t=1}^{N} \frac{B_{t}}{(log x)^{t−1/2}} + O (\frac{x}{(log x)^{N+1/2}} ) ,

valid for every real x ≥ 2 and every fixed integer N ≥ 1 where B_{t} (1 ≤ t ≤ N) are computable constants. In particular,

B_{1} = A_{1} (\frac{1}{log 2} − 1) .

Proof. Now

S_{1/τ}(x) = ∑_{ν=0}^{∞} \frac{1}{τ(2^{ν})}x^{ν} = ∑_{ν=0}^{∞} \frac{1}{ν + 1}x^{ν} = −\frac{log(1 − x)}{x} (|x| < 1)

and the reciprocal power series is

\overline{S}_{1/τ}(x) = −\frac{x}{log(1 − x)} = ∑_{ν=0}^{∞} b_{ν}x^{ν},

where b_{0} = 1, b_{1} = −1/2, b_{2} = −1/12, b_{3} = −1/24, etc. Note that the sequence (\frac{1}{ν+1})_{ν≥0} is log-convex. According to Lemma 8 (this example was considered by Kaluza [16]),

−\frac{1}{ν + 1} ≤ b_{ν} ≤ 0 (ν ≥ 1).

This shows, using (20), that

(−1)^{n−1} \frac{1}{τ(n)} = ∑_{dj=n} h_{1/τ}(d)\frac{1}{τ(j)} (n ≥ 1),

where the function h_{1/τ} is multiplicative, h_{1/τ}(2^{ν}) ≪ \frac{1}{ν} as ν → ∞ and h_{1/τ}(p^{ν}) = 0 for every prime p > 2 and ν ≥ 1.

Hence

T (x) := ∑_{n≤x}(−1)^{n−1} \frac{1}{τ(n)} = ∑_{d≤x/2} h_{1/τ}(d) ∑_{j≤x/d} \frac{1}{τ(j)} + ∑_{x/2<d≤x} h_{1/τ}(d)

= ∑_{d≤x/2} h_{1/τ}(d) (\frac{x}{d} ∑_{j=1}^{N} \frac{A_{j}}{(log(x/d))^{j−1/2}} + O (\frac{x/d}{(log(x/d))^{N+1/2}} )) + ∑_{x/2<d≤x} h_{1/τ}(d)

= x ∑_{j=1}^{N} \frac{A_{j}}{(log x)^{j−1/2}} ∑_{d≤x/2} \frac{h_{1/τ}(d)}{d(1 − \frac{log d}{log x})^{j−1/2}} + O (\frac{x}{(log x)^{N+1/2}} ∑_{d≤x/2} \frac{|h_{1/τ}(d)|}{d(1 − \frac{log d}{log x})^{N+1/2}} )

+ ∑_{x/2<d≤x} h_{1/τ}(d).

Here the last term is small:

∑_{x/2<d≤x} h_{1/τ}(d) ≪ ∑_{d=2^{ν}≤x} |h_{1/τ}(2^{ν})| ≪ ∑_{ν≤log x/ log 2} \frac{1}{ν} ≪ log log x.

Using the power series expansion

(1 + x)^{t} = ∑_{j=0}^{∞} \binom{t}{j}x^{j} (x, t ∈ ℝ, |x| < 1),

we deduce

∑_{d≤x/2} \frac{|h_{1/τ}(d)|}{d(1 − \frac{log d}{log x})^{N+1/2}} = ∑_{d≤x/2} \frac{|h_{1/τ}(d)|}{d} (1 + O (\frac{log d}{log x}))

= ∑_{d=2^{ν}≤x/2} \frac{|h_{1/τ}(2^{ν})|}{2^{ν}} + O (\frac{1}{log x} ∑_{d=2^{ν}≤x/2} \frac{|h_{1/τ}(2^{ν})|}{2^{ν}} log 2^{ν})

≪ ∑_{2^{ν}≤x/2} \frac{1}{ν2^{ν}} + \frac{1}{log x} ∑_{2^{ν}≤x/2} \frac{1}{2^{ν}} ≪ 1.

Therefore, the remainder term of above is

O (\frac{x}{(log x)^{N+1/2}} ) .

Furthermore,

∑_{d≤x/2} \frac{h_{1/τ}(d)}{d(1 − \frac{log d}{log x})^{j−1/2}} = ∑_{d≤x/2} \frac{h_{1/τ}(d)}{d} ∑_{ℓ=0}^{∞}(−1)^{ℓ}\binom{−j + 1/2}{ℓ} (\frac{log d}{log x})^{ℓ}

= ∑_{ℓ=0}^{∞}(−1)^{ℓ}\binom{−j + 1/2}{ℓ}\frac{1}{(log x)^{ℓ}} ∑_{d≤x/2} \frac{h_{1/τ}(d)}{d}(log d)^{ℓ}

= ∑_{ℓ=0}^{∞}(−1)^{ℓ}\binom{−j + 1/2}{ℓ}\frac{1}{(log x)^{ℓ}} (K_{ℓ} + O (\frac{(log x)^{ℓ−1}}{x} )) ,

where for every ℓ ≥ 0 the series

K_{ℓ} := ∑_{d=1}^{∞} \frac{h_{1/τ}(d)}{d}(log d)^{ℓ} = ∑_{\begin{matrix}d=2^{ν} \\ ν≥0 \end{matrix}} \frac{h_{1/τ}(2^{ν})}{2^{ν}}(log 2^{ν})^{ℓ}

is absolutely convergent, since |h_{1/τ}(2^{ν})| ≪ \frac{1}{ν}, and

∑_{d>x/2} \frac{|h_{1/τ}(d)|}{d}(log d)^{ℓ} = ∑_{d=2^{ν}>x/2} \frac{|h_{1/τ}(2^{ν})|}{2^{ν}}(log 2^{ν})^{ℓ} ≪ ∑_{ν>log x/ log 2} \frac{ν^{ℓ−1}}{2^{ν}} ≪ \frac{(log x)^{ℓ−1}}{x} .

We deduce that

T (x) = x ∑_{j=1}^{N} \frac{A_{j}}{(log x)^{j−1/2}} ∑_{ℓ=0}^{∞}(−1)^{ℓ}\binom{−j + 1/2}{ℓ}\frac{1}{(log x)^{ℓ}} (K_{ℓ} + O (\frac{(log x)^{ℓ−1}}{x} ))

+O (\frac{x}{(log x)^{N+1/2}} )

= x ∑_{t=1}^{N} \frac{1}{(log x)^{t−1/2}} ∑_{j=1}^{N}(−1)^{t−j}\binom{−j + 1/2}{t − j}A_{j}K_{t−j} + O (\frac{x}{(log x)^{N+1/2}} ) .

The proof is complete by denoting

B_{t} = ∑_{j=1}^{N}(−1)^{t−j}\binom{−j + 1/2}{t − j}A_{j}K_{t−j},

where B_{1} = A_{1}K_{0} = A_{1}(\frac{1}{log 2} − 1) by (21) (applied for s = 1).

Note that a similar asymptotic formula can be deduced for the alternating sum

∑_{n≤x}(−1)^{n−1} \frac{1}{τ_{k}(n)},

where τ_{k}(n) is the Piltz divisor function, based on the result for ∑_{n≤x} \frac{1}{τ_{k}(n)}, due to De Koninck and Ivić [9, Thm. 1.2].

4.5 Gcd-sum function

Let P (n) = ∑_{k=1}^{n} gcd(k, n) be the gcd-sum function. Known results include the following: P is multiplicative, P (p^{ν}) = p^{ν−1}(p(ν + 1) − ν) (ν ≥ 1),

∑_{n=1}^{∞} \frac{P (n)}{n^{s}} = \frac{ζ^{2}(s − 1)}{ζ(s)} (ℜs > 2),

∑_{n≤x} P (n) = \frac{3}{π^{2}} x^{2} (log x + 2γ − \frac{1}{2} − \frac{ζ′(2)}{ζ(2)}) + O (x^{1+θ+ε}) ,

where θ is the best exponent in Dirichlet’s divisor problem. See the survey of the author [37]. We have

Proposition 28.

∑_{n=1}^{∞}(−1)^{n−1} \frac{P (n)}{n^{s}} = (2 (1 − \frac{1}{2^{s−1}} )^{2} (1 − \frac{1}{2^{s}} )^{−1} − 1) \frac{ζ^{2}(s − 1)}{ζ(s)} (ℜs > 2).

Theorem 29.

∑_{n≤x}(−1)^{n−1}P (n) = −\frac{1}{π^{2}} x^{2} (log x + 2γ − \frac{1}{2} − \frac{ζ′(2)}{ζ(2)} − \frac{10}{3} log 2) + O (x^{1+θ+ε}) ,

where θ is the best exponent in Dirichlet’s divisor problem.

Proof. Similar to the proofs of above. Here h_{P} (2) = −6, h_{P} (2^{ν}) = 2 (ν ≥ 2).

The next formula was proved by Chen and Zhai [7, Thm. 4], sharpening a result of the author [37, Thm. 6]:

∑_{n≤x} \frac{1}{P (n)} = ∑_{j=0}^{N} \frac{K_{j}}{(log x)^{j−1/2}} + O (\frac{1}{(log x)^{N+1/2}} ) ,

valid for every real x ≥ 2 and every fixed integer N ≥ 1 where K_{j} (1 ≤ j ≤ N) are computable constants,

K_{0} = \frac{2}{\sqrt{π}} ∏_{p∈ℙ} (1 − \frac{1}{p})^{1/2} ∑_{ν=0}^{∞} \frac{1}{P (p^{ν})}.

We have

Theorem 30.

∑_{n≤x}(−1)^{n−1} \frac{1}{P (n)} = ∑_{t=0}^{N} \frac{D_{t}}{(log x)^{t−1/2}} + O (\frac{1}{(log x)^{N+1/2}} ) ,

valid for every real x ≥ 2 and every fixed integer N ≥ 1 where D_{t} (0 ≤ t ≤ N) are computable constants. In particular,

D_{0} = K_{0} (\frac{1}{2(log 2 − 1)} − 1) .

Proof. Similar to the proof of Theorem 27. Here P (2^{ν}) = (ν + 2)2^{ν−1} (ν ≥ 1). The crucial fact is that the sequence (\frac{1}{(ν+2)2^{ν−1}} )_{ν≥0} is log-convex, and therefore Lemma 8 can be used

to deduce that h_{1/P} (2^{ν}) ≪ \frac{1}{ν2^{ν}}.

4.6 Squarefree kernel

Now we move to the function κ(n) = ∏_{p|n} p (n ≥ 1), the squarefree kernel of n (radical of n). It is known that

∑_{n=1}^{∞} \frac{κ(n)}{n^{s}} = ζ(s) ∏_{p∈ℙ} (1 + \frac{p − 1}{p^{s}}) (ℜs > 2)

and for x ≥ 3,

∑_{n≤x} κ(n) = \frac{C}{2} x^{2} + O (R_{κ}(x)) ,

where C is the constant defined by (38), R_{κ}(x) = x^{3/2}δ(x) unconditionally and R_{κ}(x) = x^{7/5}ω(x) assuming the Riemann hypothesis (RH), with

δ(x) = exp (−c_{1}(log x)^{3/5}(log log x)^{−1/5}) ,  (44)

ω(x) = exp (c_{2}(log x)(log log x)^{−1}) ,  (45)

c_{1}, c_{2} being positive constants. These estimates are due (for a more general function) to Suryanarayana and Subrahmanyam [34, Cor. 4.3.5 and 4.4.5].

We have

Proposition 31.

∑_{n=1}^{∞}(−1)^{n−1} \frac{κ(n)}{n^{s}} = \frac{2^{s} − 3}{2^{s} + 1}ζ(s) ∏_{p∈ℙ} (1 + \frac{p − 1}{p^{s}}) (ℜs > 2).

Theorem 32.

∑_{n≤x}(−1)^{n−1}κ(n) = \frac{C}{10}x^{2} + O (R_{κ}(x)) ,

where C is given by (38) and R_{κ}(x) is defined above.

Proof. Here, to deduce the unconditional result, Proposition 6 cannot be applied, since the function δ(x) is not increasing. However, x^{ε}δ(x) is increasing for any ε > 0 and we obtain by (20),

∑_{n≤x}(−1)^{n−1}κ(n) = ∑_{d≤x} h_{κ}(d) (\frac{C}{2} (\frac{x}{d})^{2} + O ((\frac{x}{d})^{3/2} δ(x/d)))

= \frac{Cx^{2}}{2} ∑_{d≤x} \frac{h_{κ}(d)}{d^{2}} + O (x^{ε}δ(x)x^{3/2−ε} ∑_{d≤x} \frac{|h_{κ}(d)|}{d^{3/2−ε}}) .

Note that h_{κ}(2^{ν}) = 4(−1)^{ν} (ν ≥ 1). Hence the function h_{κ} is bounded and the result is obtained by the usual arguments.

Assuming RH, Proposition 6 can directly be applied, since ω(x) is increasing.

It is known that

K(x) := ∑_{n≤x} \frac{1}{κ(n)} = exp ((1 + o(1)) (\frac{8 log x}{log log x})^{1/2}) (x → ∞),

due to de Bruijn [5], confirming a conjecture of Erdős. In fact,

K(x) ∼ \frac{1}{2}e^{γ}F (log x)(log log x) (x → ∞),

where γ is Euler’s constant and

F (t) := \frac{6}{π^{2}} ∑_{m=1}^{∞} \frac{min(1, e^{t}/m)}{∏_{p|m}(p + 1)} (t ≥ 0),

which follows from a more precise asymptotic formula with error term, recently established by Robert and Tenenbaum [23, Thm. 4.3]. We point out that according to [23, Eq. (2.12)], there exists a sequence of polynomials (Q_{j})_{j≥1} with deg Q_{j} ≤ j (j ≥ 1) such that for any N ≥ 1,

F (t) = exp ((\frac{8t}{log t})^{1/2} (1 + ∑_{j=1}^{N} \frac{Q_{j}(log log t)}{(log t)^{j}} + O ((\frac{log log t}{log t})^{N+1}))) (t ≥ 3).

Note that

S_{1/κ}(x) = ∑_{ν=0}^{∞} \frac{1}{κ(2^{ν})}x^{ν} = \frac{2 − x}{2(1 − x)} (|x| < 1),

\overline{S}_{1/κ}(x) = \frac{2(1 − x)}{2 − x} = 1 − ∑_{ν=1}^{∞} \frac{x^{ν}}{2^{ν}} (|x| < 2),

therefore h_{1/κ}(2^{ν}) = −\frac{1}{2^{ν−1}} (ν ≥ 1) and ∑_{n=1}^{∞} h_{1/κ}(n) = −1. It follows that

K_{altern}(x) := ∑_{n≤x}(−1)^{n−1} \frac{1}{κ(n)} = K(x) − 2 ∑_{2≤2^{ν}≤x} \frac{1}{2^{ν}} K (\frac{x}{2^{ν}} ) .  (46)

Identity (46) and the deep analytic results of Robert and Tenenbaum [23] lead to the following:

Theorem 33. (Tenenbaum [36]) One has

K_{altern}(x) ∼ −K(x) (x → ∞)  (47)

and a genuine asymptotic formula with effective remainder term may be derived for K_{altern}(x).

4.7 Squarefree numbers

Now consider the squarefree numbers for which the characteristic function is μ^{2}, where μ is the Möbius function. It is well-known that

∑_{n=1}^{∞} \frac{μ^{2}(n)}{n^{s}} = \frac{ζ(s)}{ζ(2s)} (ℜs > 1)

and

∑_{n≤x} μ^{2}(n) = \frac{6}{π^{2}} x + O (R_{μ^{2}}(x)) ,

where R_{μ^{2}}(x) = x^{1/2}δ(x), with δ(x) defined by (44), unconditionally, due to Walfisz [39, Satz 1, p. 192], and R_{μ^{2}}(x) = x^{11/35+ε} (ε > 0) assuming RH, due very recently to Liu [17].

Proposition 34.

∑_{n=1}^{∞}(−1)^{n−1} \frac{μ^{2}(n)}{n^{s}} = \frac{2^{s} − 1}{2^{s} + 1} · \frac{ζ(s)}{ζ(2s)} (ℜs > 1).

Theorem 35.

∑_{n≤x}(−1)^{n−1}μ^{2}(n) = \frac{2}{π^{2}} x + O (R_{μ^{2}}(x)) .

Proof. Similar to the proof of Theorem 32. Note that here h_{μ^{2}}(2^{ν}) = 2(−1)^{ν} (ν ≥ 1). Hence the function h_{κ} is bounded.

4.8 Number of abelian groups of a given order

Let a(n) denote, as usual, the number of abelian groups of order n. This is another classical multiplicative function, investigated by several authors. It is known that

∑_{n=1}^{∞} \frac{a(n)}{n^{s}} = ∏_{k=1}^{∞} ζ(ks) (ℜs > 1),

∑_{n≤x} a(n) = C_{1}x + C_{2}x^{1/2} + C_{3}x^{1/3} + O (x^{1/4+ε}) ,  (48)

where

C_{j} = ∏_{\begin{matrix}k=1 \\ k≠j \end{matrix}}^{∞} ζ(k/j) (j = 1, 2, 3),

this best error term to date due to Robert and Sargos [22].

We have

Proposition 36.

∑_{n=1}^{∞}(−1)^{n−1} \frac{a(n)}{n^{s}} = (2 ∏_{k=1}^{∞} (1 − \frac{1}{2^{ks}} ) − 1) ∏_{k=1}^{∞} ζ(ks) (ℜs > 1).

Theorem 37.

∑_{n≤x}(−1)^{n−1}a(n) = C_{1}K_{1}x + C_{2}K_{2}x^{1/2} + C_{3}K_{3}x^{1/3} + O (x^{1/4+ε}) ,

where

K_{j} = 2 ∏_{k=1}^{∞} (1 − \frac{1}{2^{k/j}} ) − 1 (j = 1, 2, 3).

Note that K_{1} ≐ −0.422423, where the digits of ∏_{k=1}^{∞} (1 − \frac{1}{2^{k}} ) ≐ 0.288788 form the sequence \underline{A048651} in the OEIS [31].

Proof. We use the method described in Section 2.3. According to (20),

∑_{n≤x}(−1)^{n−1}a(n) = ∑_{d≤x} h_{a}(d) ∑_{j≤x/d} a(j).

Remark that by Euler’s pentagonal number theorem,

∏_{k=1}^{∞} (1 − \frac{1}{2^{ks}} ) = 1 + ∑_{j=1}^{∞}(−1)^{j} (\frac{1}{2^{(3j−1)js/2}} + \frac{1}{2^{(3j+1)js/2}} ) (|2^{s}| > 1),

which shows that h_{a}(2^{ν}) ∈ {−2, 0, 2} for every ν ≥ 1 and h_{a}(p^{ν}) = 0 for every prime p > 2 and every ν ≥ 1. Hence the function h_{a} is bounded. Now using (48), the proof can be carried out by the usual arguments.

It is known that

∑_{n≤x} \frac{1}{a(n)} = Dx + O (x^{1/2}(log x)^{−1/2}) ,

where

D = ∏_{p∈ℙ} (1 + ∑_{k=2}^{∞} (\frac{1}{P (k)} − \frac{1}{P (k − 1)}) \frac{1}{p^{k}} ) ≐ 0.752015

(sequence \underline{A084911} in the OEIS [31]), due to De Koninck and Ivić [9, Thm. 1.3]. Here P (k) denotes the number of unrestricted partitions of k (not to be confused with the gcd-sum function from Section 4.5, denoted also by P). See Nowak [20] for a more precise asymptotic formula.

It follows by (15) that the limit

lim_{x→∞} \frac{1}{x} ∑_{n≤x}(−1)^{n−1} \frac{1}{a(n)} = D (2 (1 + ∑_{ν=1}^{∞} \frac{1}{P (ν)2^{ν}} )^{−1} − 1)

exists.

To establish an asymptotic formula for

∑_{n≤x}(−1)^{n−1} \frac{1}{a(n)}  (49)

we need to estimate the coefficients of the reciprocal of the power series S_{1/a}(x) = 1 + ∑_{ν=1}^{∞} \frac{1}{P (ν)}x^{ν}. Here Lemma 8 cannot be used, since the sequence (a_{ν})_{ν≥0} with a_{0} = 1 and a_{ν} = \frac{1}{P (ν)} (ν ≥ 1) is not log-convex. However, observe that DeSalvo and Pak [10, Thm. 1.1] recently proved that the sequence (P (n)) is log-concave for n > 25, that is, (1/P (n)) is log-convex for n > 25.

Open Problem 38. Estimate the alternating sum (49).

4.9 Sum-of-unitary-divisors function

Recall that d is said to be a unitary divisor of n if d | n and gcd(d, n/d) = 1. Let σ^{∗}(n) denote, as usual, the sum of unitary divisors of n. The function σ^{∗} is multiplicative and σ^{∗}(p^{ν}) = p^{ν} + 1 (ν ≥ 1). One has

∑_{n=1}^{∞} \frac{σ^{∗}(n)}{n^{s}} = \frac{ζ(s)ζ(s − 1)}{ζ(2s − 1)} (ℜs > 2).

Furthermore,

∑_{n≤x} σ^{∗}(n) = \frac{π^{2}}{12ζ(3)}x^{2} + O (x(log x)^{5/3}) ,

established by Sitaramachandrarao and Suryanarayana [27, Eq. (1.4)].

Proposition 39.

∑_{n=1}^{∞}(−1)^{n−1} \frac{σ^{∗}(n)}{n^{s}} = (1 − \frac{6}{2^{s}} + \frac{6}{2^{2s}} ) (1 − \frac{2}{2^{2s}} )^{−1} \frac{ζ(s)ζ(s − 1)}{ζ(2s − 1)} (ℜs > 2). (50)

Theorem 40.

∑_{n≤x}(−1)^{n−1}σ^{∗}(n) = −\frac{π^{2}}{84ζ(3)}x^{2} + O (x(log x)^{5/3}) .

Proof. Apply Proposition 6. The Dirichlet series representation (50) can be used, the function h_{σ^{∗}} is bounded.

It is known that

∑_{n≤x} \frac{1}{σ^{∗}(n)} = B^{∗} log x + D^{∗} + O (x^{−1}(log x)^{5/3}(log log x)^{4/3}) ,

obtained by Sita Ramaiah and Suryanarayana [30, p. 1352], where B^{∗} and D^{∗} are explicit constants. Here, according to Proposition 4,

B^{∗} = ∏_{p∈ℙ} (1 − \frac{1}{p}) (1 + ∑_{ν=1}^{∞} \frac{1}{p^{ν} + 1}) .

It follows from the same Proposition 4 that the limit

E^{∗} := lim_{x→∞} \frac{1}{log x} ∑_{n≤x}(−1)^{n−1} \frac{1}{σ^{∗}(n)} = B^{∗} (2 (1 + ∑_{ν=1}^{∞} \frac{1}{2^{ν} + 1})^{−1} − 1)

exists.

Moreover, by Corollary 10 we deduce that

∑_{n≤x}(−1)^{n−1} \frac{1}{σ^{∗}(n)} = E^{∗} log x + F^{∗} + O (x^{−u}(log x)^{5/3}(log log x)^{4/3}) ,  (51)

with an explicit constant F^{∗} and some u > 0. Bordellès and Cloitre [4, Cor. 4, (vi)] established that the error term of (51) is O (x^{−1}(log x)^{4}).

To use our method, we need a better estimate for the coefficients b_{ν} of the reciprocal of the power series

S_{1/σ^{∗}}(x) = 1 + ∑_{ν=1}^{∞} \frac{x^{ν}}{2^{ν} + 1}.

Here b_{0} = 1, b_{1} = −\frac{1}{3}, b_{2} = −\frac{4}{45}, b_{3} = −\frac{2}{135}, b_{4} = \frac{32}{34425}, etc.

Open Problem 41. We conjecture that b_{ν} ≪ 1/2^{ν} as ν → ∞. If this is true, then it follows from Proposition 7 that the error term in (51) can be improved into O (x^{−1}(log x)^{8/3}(log log x)^{4/3}).

We pose as an open problem to prove this.

4.10 Unitary Euler function

Let φ^{∗} be the unitary analog of Euler’s φ function. The function φ^{∗} is multiplicative and φ(p^{ν}) = p^{ν} − 1 for every prime power p^{ν} (ν ≥ 1). One has

∑_{n=1}^{∞} \frac{φ^{∗}(n)}{n^{s}} = ζ(s)ζ(s − 1) ∏_{p} (1 − \frac{2}{p^{s}} + \frac{1}{p^{2s−1}} ) (ℜs > 2).

Furthermore,

∑_{n≤x} φ^{∗}(n) = \frac{C}{2} x^{2} + O (x(log x)^{5/3}(log log x)^{4/3}) ,

where C is defined by (38). See Sitaramachandrarao and Suryanarayana [27, Eq. (1.5)].

Proposition 42.

∑_{n=1}^{∞}(−1)^{n−1} \frac{φ^{∗}(n)}{n^{s}} = (1 − \frac{1}{2^{s−2}} + \frac{1}{2^{2s−1}} ) (1 − \frac{1}{2^{s−1}} + \frac{1}{2^{2s−1}} )^{−1} ∑_{n=1}^{∞} \frac{φ^{∗}(n)}{n^{s}} (ℜs > 2),

(52)

Theorem 43.

∑_{n≤x}(−1)^{n−1}φ^{∗}(n) = \frac{C}{10}x^{2} + O (x(log x)^{5/3}(log log x)^{4/3}) ,

where C is defined by (38).

Proof. Apply Proposition 6. The Dirichlet series representation (52) can be used. The function h_{φ^{∗}} is bounded.

It is known that

∑_{n≤x} \frac{1}{φ^{∗}(n)} = L^{∗} log x + M^{∗} + O (x^{−1}(log x)^{5/3}) ,

due to Sita Ramaiah and Subbarao [28, Thm. 3.1], improving the error term of Sita Ramaiah and Suryanarayana [30, Thm. 3.2], where L^{∗} and M^{∗} are explicit constants. Here, according to Proposition 4,

L^{∗} = ∏_{p∈ℙ} (1 − \frac{1}{p}) (1 + ∑_{ν=1}^{∞} \frac{1}{p^{ν} − 1}) .

It follows from the same Proposition 4 that the limit

T^{∗} := lim_{x→∞} \frac{1}{log x} ∑_{n≤x}(−1)^{n−1} \frac{1}{φ^{∗}(n)} = L^{∗} (\frac{2}{1 + K} − 1)

exists, where K is the Erdős-Borwein constant defined by (43). Moreover, by Corollary 10 (take q = 1/2) we deduce that

∑_{n≤x}(−1)^{n−1} \frac{1}{φ^{∗}(n)} = T^{∗} log x + U^{∗} + O (x^{−u}(log x)^{5/3}) .  (53)

with an explicit constant U^{∗} and some u > 0.

Note that this example was not considered by Bordellès and Cloitre [4]. To use our method one needs to consider the power series

S_{1/φ^{∗}}(x) = 1 + ∑_{ν=1}^{∞} \frac{x^{ν}}{2^{ν} − 1},

where the sequence a_{0} = 1, a_{ν} = \frac{1}{2^{ν}−1} is log-convex but only for ν ≥ 1, that is a_{ν}^{2} ≤ a_{ν−1}a_{ν+1} holds for ν ≥ 2 and is false for ν = 1. Hence Lemma 8 cannot be used. In fact, the coefficients b_{ν} of the reciprocal power series are b_{0} = 1, b_{1} = −1, b_{2} = \frac{2}{3}, b_{3} = −\frac{10}{21}, b_{4} = \frac{104}{315}, etc. (not all of b_{1}, b_{2}, … are negative).

Open Problem 44. We conjecture that b_{ν} ≪ 1/2^{ν} as ν → ∞. If this is true, then it follows from Proposition 7 that the error term in (53) can be improved into O (x^{−1}(log x)^{8/3}).

We pose as an open problem to prove this.

4.11 Unitary squarefree kernel

Let κ^{∗}(n) denote the greatest squarefree unitary divisor of n. The function κ^{∗} is multiplicative, κ^{∗}(p) = p and κ^{∗}(p^{ν}) = 1 for every prime p and ν ≥ 2. One has

∑_{n≤x} κ^{∗}(n) = \frac{1}{2}C̃x^{2} + O(R_{κ^{∗}}(x)),  (54)

where

C̃ = ∏_{p∈ℙ} (1 − \frac{p^{2} + p − 1}{p^{3}(p + 1)}) ,  (55)

R_{κ^{∗}}(x) = x^{3/2}δ(x), with δ(x) defined by (44), unconditionally, and R_{κ^{∗}}(x) = x^{7/5}ω(x), with ω(x) defined by (45), assuming RH, due to Sita Ramaiah and Suryanarayana [26, Thm. 5.7, 5.8].

Theorem 45. With the notation above,

∑_{n≤x}(−1)^{n−1}κ^{∗}(n) = \frac{5}{38}C̃x^{2} + O(R_{κ^{∗}}(x)).  (56)

Proof. We have

S_{κ^{∗}}(x) = \frac{x^{2} − x − 1}{x − 1} (|x| < 1)

and the proof is quite similar to the proof of Theorem 32.

An asymptotic formula for the reciprocal of κ^{∗}(n) is simpler to obtain than for the reciprocal of the squarefree kernel κ(n), discussed in Section 4.6. It is a result of Suryanarayana and Subrahmanyam [33, Cor. 3.4.1] that

∑_{n≤x} \frac{1}{κ^{∗}(n)} = \frac{Aζ(3/2)}{ζ(3)}x^{1/2} + \frac{Bζ(2/3)}{ζ(2)}x^{1/3} + O(x^{1/5}),  (57)

where

A = ∏_{p∈ℙ} (1 + \frac{\sqrt{p} − 1}{p(p − \sqrt{p} + 1)}) , B = ∏_{p∈ℙ} (1 + \frac{p^{1/3} − 1}{p(p^{2/3} − p^{1/3} + 1)}) ,

We deduce the next result.

Theorem 46.

∑_{n≤x}(−1)^{n−1} \frac{1}{κ^{∗}(n)} = \frac{A^{∗}ζ(3/2)}{ζ(3)}x^{1/2} + \frac{B^{∗}ζ(2/3)}{ζ(2)}x^{1/3} + O(x^{1/5}),  (58)

where

A^{∗} = \frac{A(9 − 12\sqrt{2})}{23}, B^{∗} = \frac{B(2^{5/3} − 3 · 2^{1/3} − 1)}{2^{5/3} − 2^{1/3} + 1}.  (59)

Proof. We have

S_{1/κ^{∗}}(x) = \frac{x^{2} − x + 2}{2(1 − x)} (|x| < 1),

hence

\overline{S}_{1/κ^{∗}}(x) = \frac{2(1 − x)}{x^{2} − x + 2} = ∑_{ν=0}^{∞} b_{ν}x^{ν} (|x| < \sqrt{2}),

where

b_{ν} = \frac{1}{4^{ν+1}\sqrt{7}} Re ((\sqrt{7} + i)(1 − i\sqrt{7})^{ν+1} + (\sqrt{7} − i)(1 + i\sqrt{7})^{ν+1}) (ν ≥ 0).

Therefore,

|b_{ν}| ≤ \frac{4}{\sqrt{7}} · \frac{1}{2^{ν/2}} (ν ≥ 1)

and using our method this implies (58).

4.12 Powerful part of a number

It is possible to deduce similar formulas for many other special multiplicative functions. We consider the following further example. Every positive integer n can be uniquely written as n = ab, where gcd(a, b) = 1, a is squarefree and b is squareful. Here b is called the powerful part of n and is denoted by pow(n). See Cloutier, De Koninck, Doyon [8]. Note that

pow(n) = \frac{n}{κ^{∗}(n)} (n ≥ 1),  (60)

where κ^{∗}(n) is the unitary squarefree kernel of n, discussed in Section 4.11.

By partial summation we deduce from (57) that

∑_{n≤x} pow(n) = \frac{1}{3}c_{1}x^{3/2} + \frac{1}{4}c_{2}x^{4/3} + O(x^{6/5}),  (61)

where

c_{1} = ∏_{p∈ℙ} (1 + \frac{2}{p^{3/2}} − \frac{1}{p^{5/2}} ) ,

c_{2} = ∏_{p∈ℙ} (1 + \frac{1}{p^{2/3}} + \frac{2}{p^{4/3}} − \frac{1}{p^{7/3}} ) .

We remark that (61) is better than [8, Eq. (1)], where the error term is O(x^{4/3}).

Theorem 47.

∑_{n≤x}(−1)^{n−1} pow(n) = \frac{A^{∗}ζ(3/2)}{3ζ(3)}x^{3/2} + \frac{B^{∗}ζ(4/3)}{4ζ(2)}x^{4/3} + O(x^{6/5}),

where the constants A^{∗} and B^{∗} are defined by (59).

Proof. Use formulas (60), (58) and partial summation. Alternatively, formula (61) and the method of the present paper can be applied.

By partial summation again, we deduce from (60) and (54) that

∑_{n≤x} \frac{1}{pow(n)} = C̃x + O(R_{1/ pow}(x))  (62)

where C̃ is defined by (55), R_{1/ pow}(x) = x^{1/2}δ(x), with δ(x) defined by (44), unconditionally, and R_{1/ pow}(x) = x^{2/5}ω(x), with ω(x) defined by (45), assuming RH. Note that this error term is better than O(x^{1/2}), indicated in [8, Eq. (3)].

Theorem 48.

∑_{n≤x}(−1)^{n−1} \frac{1}{pow(n)} = \frac{5}{19}C̃x + O(R_{1/ pow}(x)),

with the notation above.

Proof. Apply formulas (60), (56) and partial summation. Alternatively, formula (62) and the method of the present paper can be used.

4.13 Sum-of-bi-unitary-divisors function

Let σ^{∗∗}(n) denote, as usual, the sum of bi-unitary divisors of n. Recall that a divisor d of n is a bi-unitary divisor if the greatest common unitary divisor of d and n/d is 1. The function σ^{∗∗} is multiplicative and for any prime power p^{ν} (ν ≥ 1),

σ^{∗∗}(p^{ν}) = \begin{cases}σ(p^{ν}), & if ν is odd; \\ σ(p^{ν}) − p^{ν/2}, & if ν is even. \end{cases}

It is the result of Suryanarayana and Subbarao [32, Cor. 3.4.3] that

∑_{n≤x} σ^{∗∗}(n) = \frac{1}{2}C^{∗∗}x^{2} + O(x(log x)^{3}),

where

C^{∗∗} = ζ(2)ζ(3) ∏_{p∈ℙ} (1 − \frac{2}{p^{3}} + \frac{1}{p^{4}} + \frac{1}{p^{5}} − \frac{1}{p^{6}} ) .

Theorem 49. We have

∑_{n≤x}(−1)^{n−1}σ^{∗∗}(n) = −\frac{11}{106}C^{∗∗}x^{2} + O(x(log x)^{3}).

Proof. Similar to the proof of (40), by applying Proposition 6 for f = σ^{∗∗}.

Sitaramaiah and Subbarao [28, Thm. 3.2] established that

∑_{n≤x} \frac{1}{σ^{∗∗}(n)} = A^{∗∗} log x + B^{∗∗} + O(x^{−1}(log x)^{14/3}(log log x)^{4/3}),

where A^{∗∗}, B^{∗∗} are certain explicit constants.

Theorem 50. We have

∑_{n≤x}(−1)^{n−1} \frac{1}{σ^{∗∗}(n)} = A_{1}^{∗∗} log x + B_{1}^{∗∗} + O(x^{c}(log x)^{14/3}(log log x)^{4/3}),  (63)

where A_{1}^{∗∗}, B_{1}^{∗∗} are explicit constants and c = (log 9/10)/(log 2) ≐ −0.152003.

Proof. Now Lemma 8 (theorem of Kaluza) cannot be used, since the sequence (\frac{1}{σ^{∗∗}(2^{ν})})_{ν≥0}

is not log-convex. But it is easy to check that

\frac{1}{σ^{∗∗}(2^{ν})} ≤ \frac{4}{5} · \frac{1}{2^{ν}} (ν ≥ 1),

hence Corollary 13 can be applied with A = 4/5, q = 1/2, where M = q(A + 1) = 9/10 < 1.

Open Problem 51. Improve the error term of (63).

4.14 Alternating sum-of-divisors function

Consider the function β(n) = ∑_{d|n} d λ(n/d) (n ≥ 1), where λ is the Liouville function. The function β is multiplicative and β(p^{ν}) = p^{ν} − p^{ν−1} + p^{ν−2} − ⋯ + (−1)^{ν} for every prime power p^{ν} (ν ≥ 1). See the survey paper of the author [38].

Proposition 52.

∑_{n=1}^{∞}(−1)^{n−1} \frac{β(n)}{n^{s}} = (1 − \frac{2}{2^{s}} − \frac{4}{2^{2s}} ) \frac{ζ(s − 1)ζ(2s)}{ζ(s)} (ℜs > 2).  (64)

Proof. Use Proposition 1 and the representation

∑_{n=1}^{∞} \frac{β(n)}{n^{s}} = \frac{ζ(s − 1)ζ(2s)}{ζ(s)} (ℜs > 2).

Theorem 53.

∑_{n≤x}(−1)^{n−1}β(n) = \frac{π^{2}}{120}x^{2} + O (x(log x)^{2/3}(log log x)^{4/3}) .  (65)

Proof. Apply Proposition 6 for f = β. It is known that

∑_{n≤x} β(n) = \frac{π^{2}}{30}x^{2} + O (x(log x)^{2/3}(log log x)^{4/3}) ,

see [38, Eq. (15)], which is a consequence of the result of Walfisz [39, Satz 4, p. 144] for Euler’s φ function. The coefficient of the main term in (65) is from (64),

\frac{π^{2}}{30} [1 − \frac{2}{2^{s}} − \frac{4}{2^{2s}} ]_{s=2} = \frac{π^{2}}{120}.

It is known ([38, Eq. (17)]) that for every ε > 0,

∑_{n≤x} \frac{1}{β(n)} = K_{1} log x + K_{2} + O(x^{−1+ε}),  (66)

where K_{1} and K_{2} are constants. Since (β(2^{ν}))_{ν≥0} is nondecreasing and β(2^{ν}) ≥ 2^{ν−1} (ν ≥ 1), Corollary 10 can be applied (take q = 1/2). We deduce that

∑_{n≤x}(−1)^{n−1} \frac{1}{β(n)} = K_{3} log x + K_{4} + O(x^{−u})  (67)

with some constants K_{3}, K_{4} and some u > 0.

Open Problem 54. Improve the error terms of (66) and (67).

4.15 Exponential divisor function

The exponential divisor function τ^{(e)} is multiplicative and τ^{(e)}(p^{ν}) = τ(ν) for every prime power p^{ν} (ν ≥ 1), where τ is the classical divisor function. There are constants A_{1} and A_{2} such that

∑_{n≤x} τ^{(e)}(n) = A_{1}x + A_{2}x^{1/2} + O(R_{τ^{(e)}}(x)),

where

A_{1} = ∏_{p∈ℙ} (1 + ∑_{ν=2}^{∞} \frac{τ(ν) − τ(ν − 1)}{p^{ν}})

is the mean value of τ^{(e)} and R_{τ^{(e)}}(x) = x^{2/9} log x, as shown by Wu [41, Thm. 1]. This error term is strongly related to estimates on the divisor function d(1, 2; n) = ∑_{ab^{2}=n} 1. It can be sharpened into O(x^{1057/4785+ε}) by using [13, Thm. 1]. Also see [41, Remark, p. 135].

It follows from Proposition 2 that the limit

lim_{x→∞} \frac{1}{x} ∑_{n≤x}(−1)^{n−1}τ^{(e)}(n) = A_{1} (\frac{2}{1 + K} − 1)

exists, where K = ∑_{ν=1}^{∞} \frac{τ(ν)}{2^{ν}} = ∑_{ν=1}^{∞} \frac{1}{2^{ν}−1} is the Erdős-Borwein constant, already quoted above.

Open Problem 55. Investigate the alternating sums

∑_{n≤x}(−1)^{n−1}τ^{(e)}(n),  ∑_{n≤x}(−1)^{n−1} \frac{1}{τ^{(e)}(n)}.

5 Generalized alternating sums

It is possible to investigate the following generalization of the alternating sums discussed above. Let Q be an arbitrary subset of the set of primes ℙ, let

t_{Q}(n) := \begin{cases}1, & if q ∤ n for every q ∈ Q; \\ −1, & otherwise, \end{cases}

and

D_{Q}(f, s) := ∑_{n=1}^{∞} t_{Q}(n)\frac{f(n)}{n^{s}}.  (68)

If Q = {2}, then t_{{2}}(n) = (−1)^{n−1} and (68) reduces to the alternating Dirichlet series (4). If Q = {2, 3}, just to illustrate another special case, then we have

D_{\{2,3\}}(f, s) = \frac{f(1)}{1^{s}} − \frac{f(2)}{2^{s}} − \frac{f(3)}{3^{s}} − \frac{f(4)}{4^{s}} + \frac{f(5)}{5^{s}} − \frac{f(6)}{6^{s}} + \frac{f(7)}{7^{s}} + ⋯,

while the choice Q = ∅ gives the classical Dirichlet series (5).

Note that the function n ↦ t_{Q}(n) is multiplicative if and only if Q = {q} having one element. Proof: If Q = {q}, then the function t_{{q}}(n) is multiplicative. On the other hand, if there are distinct q_{1}, q_{2} ∈ Q, then t_{Q}(q_{1}q_{2}) = −1 ≠ 1 = (−1)(−1) = t_{Q}(q_{1})t_{Q}(q_{2}). However, the function

c_{Q}(n) := \begin{cases}1, & if q ∤ n for every q ∈ Q; \\ 0, & otherwise; \end{cases}

is multiplicative for every Q ⊆ ℙ.

Proposition 56. Let Q be an arbitrary subset of ℙ. If f is a multiplicative function, then

D_{Q}(f, s) = D(f, s) (2 ∏_{q∈Q} (∑_{ν=0}^{∞} \frac{f(q^{ν})}{q^{νs}})^{−1} − 1) ,

and if f is completely multiplicative, then

D_{Q}(f, s) = ∏_{p∈ℙ} (1 − \frac{f(p)}{p^{s}})^{−1} (2 ∏_{q∈Q} (1 − \frac{f(q)}{q^{s}}) − 1) ,

formally or in case of convergence.

Proof. We have

D_{Q}(f, s) = − ∑_{n=1}^{∞} \frac{f(n)}{n^{s}} + 2 ∑_{n=1}^{∞} c_{Q}(n)\frac{f(n)}{n^{s}} = −D(f, s) + 2 ∏_{p∉Q} ∑_{ν=0}^{∞} \frac{f(p^{ν})}{p^{νs}}

= −D(f, s) + 2D(f, s) ∏_{q∈Q} (∑_{ν=0}^{∞} \frac{f(q^{ν})}{q^{νs}})^{−1} = D(f, s) (2 ∏_{q∈Q} (∑_{ν=0}^{∞} \frac{f(q^{ν})}{q^{νs}})^{−1} − 1) .

If Q = {2}, then Proposition 56 reduces to Proposition 1.

Some of the discussed asymptotic formulas can also be generalized to certain subsets Q ⊆ ℙ. For example, we have the next result.

Theorem 57. Let Q be an arbitrary finite subset of ℙ. Then

∑_{n≤x} t_{Q}(n)σ(n) = \frac{π^{2}}{12} (2 ∏_{p∈Q} (1 − \frac{1}{p}) (1 − \frac{1}{p^{2}} ) − 1) x^{2} + O (x(log x)^{2/3}) .  (69)

Proof. We have

∑_{n≤x} t_{Q}(n)σ(n) = − ∑_{n≤x} σ(n) + 2 ∑_{n≤x} c_{Q}(n)σ(n),

where c_{Q}(n)σ(n) is multiplicative and

∑_{n=1}^{∞} \frac{c_{Q}(n)σ(n)}{n^{s}} = ζ(s)ζ(s − 1) ∏_{p∈Q} (1 − \frac{p + 1}{p^{s}} + \frac{p}{p^{2s}} ) .

It turns out that

∑_{n≤x} c_{Q}(n)σ(n) = ∑_{d≤x} h_{Q}(d) ∑_{j≤x/d} σ(j),

where the function h_{Q} is multiplicative and for every prime power p^{ν} (ν ≥ 1),

h_{Q}(p^{ν}) = \begin{cases}−(p + 1), & if p ∈ Q, ν = 1; \\ p, & if p ∈ Q, ν = 2; \\ 0, & otherwise. \end{cases}

Now the proof runs similar to the proof of (40).

In the case Q = {2} formula (69) reduces to (40).

Open Problem 58. Deduce asymptotic formulas for

∑_{n≤x} t_{Q}(n)σ(n)

and for similar sums if Q is an arbitrary fixed subset of the primes.

6 Acknowledgments

The author thanks Shyamal Biswas for raising the question of investigating alternating Dirichlet series, Roberto Tauraso for drawing attention to the paper of Kaluza [16], and Gérald Tenenbaum for helpful remarks on Section 4.6. The author is grateful to the anonymous referee for identity (27), Proposition 12 and many other useful comments and suggestions.

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2010 Mathematics Subject Classification: Primary 11N37; Secondary 11A05, 11A25, 30B10. Keywords: multiplicative arithmetic function, alternating sum, Dirichlet series, asymptotic formula, reciprocal power series, Euler’s totient function, Dedekind function, sum-of-divisors function, divisor function, gcd-sum function, unitary divisor.

(Concerned with sequences \underline{A000005}, \underline{A000010}, \underline{A000041}, \underline{A000203}, \underline{A000688}, \underline{A001615}, \underline{A002088}, \underline{A005117}, \underline{A006218}, \underline{A007947}, \underline{A013928}, \underline{A018804}, \underline{A024916}, \underline{A033999}, \underline{A034448}, \underline{A047994}, \underline{A048651}, \underline{A049419}, \underline{A057521}, \underline{A063966}, \underline{A064609}, \underline{A065442}, \underline{A065463}, \underline{A068762}, \underline{A068773}, \underline{A084911}, \underline{A143348}, \underline{A145353}, \underline{A173290}, \underline{A177754}, \underline{A188999}, \underline{A206369}, \underline{A272718}.)

Received August 4 2016; revised versions received October 20 2016; December 13 2016. Published in Journal of Integer Sequences, December 27 2016.

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