An Alternating Sum Involving the Reciprocal of Certain Multiplicative Functions

Olivier Bordellès 2, allée de la Combe 43000 Aiguilhe France

                  borde43@wanadoo.fr

Benoit Cloitre 19, rue Louise Michel 92300 Levallois-Perret France

                 benoit7848c@yahoo.fr

To the memory of Patrick Sargos

Abstract

We establish an asymptotic formula for an alternating sum of the reciprocal of a class of multiplicative functions. The proof is straightforward and uses classical convolution techniques. Numerous examples are given.

1 Introduction and main results

In 1900, E. Landau [2] proved that

∑_{n≤x} \frac{1}{φ(n)} = \frac{ζ(2)ζ(3)}{ζ(6)} (log x + γ − ∑_{p} \frac{log p}{p^{2} − p + 1}) + O (\frac{log x}{x})

where φ is the classic Euler totient function and γ is Euler’s constant, but it seems that in the usual literature there is not any similar result for the alternating sum

∑_{n≤x} \frac{(−1)^{n}}{φ(n)} .

It is fairly easy to show that there exists an absolute constant c > 0 such that this sum is > c log x + O(1), and therefore the question of an asymptotic formula arises naturally. It should be mentioned that Landau’s result could eventually provide such an asymptotic formula via the simple identity

∑_{n≤x} \frac{(−1)^{n}}{φ(n)} = ∑_{n≤x} \frac{1}{φ(n)} − 2 ∑_{\begin{matrix}n≤x \\ (n,2)=1 \end{matrix}} \frac{1}{φ(n)}

and the use of known estimates as in [3, Theorem 1], but this method does not seem to be easily generalized to a larger class of multiplicative functions than that of [3]. In this article, we will follow a slightly different approach working with the set of non-zero complex-valued multiplicative functions f satisfying the following assumptions: there exist constants λ_{1} > 0 and 0 ≤ λ_{2} < 2 such that, for each prime power p^{α}, we have

p^{2} |\frac{1}{f(p)} − \frac{1}{p}| ≤ λ_{1} and p^{2α−1} |\frac{1}{f (p^{α})} − \frac{1}{pf (p^{α−1})}| ≤ λ_{1}λ_{2}^{α−1} (α ≥ 2) .  (1)

It will be convenient to set

λ := \frac{λ_{1}(2 + λ_{2})}{2 − λ_{2}}.  (2)

For any prime number p, define

S_{f}(p) := ∑_{α=1}^{∞} \frac{1}{f(p^{α})} and s_{f}(p) := ∑_{α=1}^{∞} \frac{α}{f(p^{α})}  (3)

and set

P_{f} := ∏_{p} (1 − \frac{1}{p}) (1 + ∑_{α=1}^{∞} \frac{1}{f(p^{α})})  (4)

where the product is absolutely convergent.

We are now in a position to state our main result.

Theorem 1. Let f be a non-zero multiplicative function satisfying (1). Then, for x ≥ e, we have

∑_{n≤x} \frac{(−1)^{n}}{f(n)} = \frac{S_{f}(2) − 1}{S_{f}(2) + 1} P_{f} log x + C_{f} + O (\frac{(log x)^{λ+1}}{x} )

where

C_{f} := \frac{S_{f}(2) − 1}{S_{f}(2) + 1} P_{f} (γ + ∑_{p} log p (\frac{1}{p − 1} − \frac{s_{f}(p)}{1 + S_{f}(p)}) − \frac{2s_{f}(2) log 2}{S_{f}(2)^{2} − 1})

and S_{f}(p), s_{f}(p) and P_{f} are given in (3) and (4).

Corollary 2. Let f be a non-zero multiplicative function satisfying (1) and assume that, for any prime p and any integer α ≥ 1, f (p^{α}) = p^{α−1}f(p). Then, for x ≥ e, we have

∑_{n≤x} \frac{(−1)^{n}}{f(n)} = \frac{2 − f(2)}{2 + f(2)} ∏_{p} (1 − \frac{1}{p} + \frac{1}{f(p)}) (log x + γ + ∑_{p} \frac{(f(p) − p) log p}{(p − 1)f(p) + p} − \frac{8f(2) log 2}{4 − f(2)^{2}})

+ O (\frac{(log x)^{λ_{1}+1}}{x} ) .

Remark 3. If S_{f}(2) = 1 in Theorem 1, i.e. f(2) = 2 in Corollary 2, then the constant C_{f} reduces to

C_{f} = −P_{f} s_{f}(2) log \sqrt{2}.

Corollary 4. For x ≥ e, the following estimates hold.

(i)

∑_{n≤x} \frac{(−1)^{n}}{φ(n)} = \frac{ζ(2)ζ(3)}{3ζ(6)} (log x + γ − ∑_{p} \frac{log p}{p^{2} − p + 1} − \frac{8 log 2}{3}) + O (\frac{(log x)^{3}}{x} ) .

(ii) Let χ be a non-principal Dirichlet character modulo q > 2 and φ(n, χ) be the twisted Euler function attached to χ given in (6). Then

∑_{n≤x} \frac{(−1)^{n}}{φ(n, χ)} = \frac{χ(2)L(1, χ)}{4 − χ(2)} ∏_{p} (1 − \frac{χ(p)}{p} + \frac{χ(p)}{p^{2}}) (log x + γ

− ∑_{p} \frac{χ(p) log p}{p^{2} − pχ(p) + χ(p)} − \frac{8(2 − χ(2)) log 2}{χ(2)(4 − χ(2))}) + O (\frac{(log x)^{3}}{x} ) .

(iii) Let Ψ be the Dedekind arithmetic function defined in (5). Then

∑_{n≤x} \frac{(−1)^{n}}{Ψ(n)} = −\frac{1}{5} ∏_{p} (1 − \frac{1}{p(p + 1)}) (log x + γ + ∑_{p} \frac{log p}{p^{2} + p − 1} + \frac{24 log 2}{5})

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

(iv) Let γ(n) := ∏_{p|n} p be squarefree kernel of n. Then

∑_{n≤x} \frac{(−1)^{n}γ(n)}{φ(n)^{2}} = \frac{5}{11} ∏_{p} (1 + \frac{p^{2} + p − 1}{p(p + 1)(p − 1)^{2}} ) (log x + γ

− ∑_{p} \frac{(p^{3} − 2p^{2} − p + 1) log p}{(p^{2} − 1)(p^{4} − p^{3} + 2p − 1)} − \frac{64 log 2}{55}) + O (\frac{(log x)^{7}}{x} ) .

(v) Let σ be the sum-of-divisors function. Then

∑_{n≤x} \frac{(−1)^{n}}{σ(n)} = \frac{S_{σ}(2) − 1}{S_{σ}(2) + 1} ∏_{p} (1 − \frac{1}{p} + \frac{(p − 1)^{2}}{p} ∑_{α=2}^{∞} \frac{1}{p^{α} − 1}) (log x + γ

+ ∑_{p} log p (\frac{1}{p − 1} − \frac{s_{σ}(p)}{1 + S_{σ}(p)}) + κ log 2) + O (\frac{(log x)^{4}}{x} )

where κ ≐ 3.6. Note that the leading constant is ≐ −0.16 468.

(vi) Let σ^{∗} be the sum-of-unitary-divisors function. Then

∑_{n≤x} \frac{(−1)^{n}}{σ^{∗}(n)} = \frac{S_{σ^{∗}}(2) − 1}{S_{σ^{∗}}(2) + 1} ∏_{p} (1 − \frac{1}{p} + \frac{p − 1}{p} ∑_{α=1}^{∞} \frac{1}{p^{α} + 1}) (log x + γ

+ ∑_{p} log p (\frac{1}{p − 1} − \frac{s_{σ^{∗}}(p)}{1 + S_{σ^{∗}}(p)}) + ν log 2) + O (\frac{(log x)^{4}}{x} )

where ν ≐ 8.04. Note that the leading constant is ≐ −0.10 259.

2 Notation

The celebrated Möbius function is as always denoted by μ, Id(n) = n, φ is the Euler totient function, Ψ is the Dedekind function defined by

Ψ(n) = n ∏_{p|n} (1 + \frac{1}{p})  (5)

and, for each fixed non-principal Dirichlet character χ modulo q > 2, the χ-twisted Euler function φ(n, χ) recently introduced in [1] is defined by

φ(n, χ) = n ∏_{p|n} (1 − \frac{χ(p)}{p}) .  (6)

For any two arithmetic functions u and v, u ⋆ v is the usual Dirichlet convolution product of u and v defined by

(u ⋆ v)(n) = ∑_{d|n} u(d)v(n/d).

The Eratosthenes transform of u is the arithmetic function u ⋆ μ. Finally, f is a non-zero multiplicative function satisfying (1) and g is the Eratosthenes transform of Id f^{−1}. Note that the assumption (1) can be written as

p|g(p)| ≤ λ_{1} and p^{α−1} |g (p^{α})| ≤ λ_{1}λ_{2}^{α−1} (α ≥ 2) .  (7)

3 Tools

Lemma 5. Let δ > 0. The Dirichlet series of the arithmetic function g(n) is absolutely convergent in the half-plane σ > δ.

Proof. Let s = δ + it ∈ ℂ with δ > 0. The function g is multiplicative and using (7) we get for all z ≥ e

∑_{p≤z} ∑_{α=1}^{∞} |\frac{g (p^{α})}{p^{sα}} | = ∑_{p≤z} (\frac{|g(p)|}{p^{δ}} + ∑_{α=2}^{∞} |\frac{g (p^{α})}{p^{sα}} |)

≤ λ_{1} (∑_{p≤z} \frac{1}{p^{δ+1}} + ∑_{p≤z} \frac{1}{p^{δ}} ∑_{α=2}^{∞} (\frac{λ_{2}}{p^{δ+1}} )^{α−1})

= λ_{1} ∑_{p≤z} (\frac{1}{p^{δ+1}} + \frac{λ_{2}}{p^{δ}(p^{δ+1} − λ_{2})})

≤ λ_{1} ∑_{p≤z} (\frac{1}{p^{δ+1}} + \frac{2λ_{2}}{p^{δ}(2 − λ_{2})} \frac{1}{p^{δ+1}} )

≤ λ ∑_{p≤z} \frac{1}{p^{δ+1}}

where we used the inequality

\frac{1}{p^{θ} − λ_{2}} ≤ \frac{2}{2 − λ_{2}} \frac{1}{p^{θ}} (θ ≥ 1) .

This implies the asserted result.

Lemma 6. For all real numbers z ≥ e and a ∈ {0, 1}

(i) : ∑_{n≤z} |g(n)| ≪ (log z)^{λ} .

(ii) : ∑_{n>z} \frac{|g(n)|(log n)^{a}}{n} ≪ z^{−1} (log z)^{λ+a} .

Proof.

(i) As in the proof of Lemma 5, we get for all z ≥ e

∑_{n≤z} |g(n)| ≤ exp {∑_{p≤z} |g(p)| + ∑_{p≤z} ∑_{α=2}^{∞} |g (p^{α})|}
≤ exp \{λ_{1} (∑_{p≤z} \frac{1}{p} + ∑_{p≤z} ∑_{α=2}^{∞} (\frac{λ_{2}}{p} )^{α−1})\}
≤ exp \{λ_{1} (∑_{p≤z} \frac{1}{p} + \frac{2λ_{2}}{2 − λ_{2}} ∑_{p≤z} \frac{1}{p})\}
= exp (λ ∑_{p≤z} \frac{1}{p}) ≪ (log z)^{λ}

as asserted. (ii) Follows from Lemma 5, (i) and partial summation. We leave the details to the reader.

The proof is complete. Lemma 7. For any real number x ≥ 1 and any integer 1 ≤ d ≤ 2x, we have

∑_{\begin{matrix}n≤x \\ d|2n \end{matrix}} \frac{1}{n} = \frac{ρ(d)}{d} (log \frac{ρ(d)x}{d} + γ) + O (\frac{1}{x})

where

ρ(d) := \begin{cases}2, & if d is even; \\ 1, & if d is odd. \end{cases}  (8)

Proof. Let S(x, d) be the sum of the left-hand side. If x < d ≤ 2x, then

S(x, d) = \frac{2}{d} < \frac{2}{x}.

If d ≤ x is odd, then by Gauss’s theorem we have

S(x, d) = ∑_{\begin{matrix}n≤x \\ d|n \end{matrix}} \frac{1}{n} = \frac{1}{d} ∑_{1≤k≤x/d} \frac{1}{k} = \frac{1}{d} (log \frac{x}{d} + γ + O (\frac{d}{x}))

as required.

Now assume that d is even. If \frac{2x}{3} < d ≤ x, then \frac{x}{d} − \frac{1}{2} < 1 and therefore

S(x, d) = ∑_{\begin{matrix}n≤x \\ d|n \end{matrix}} \frac{1}{n} + ∑_{\begin{matrix}n≤x \\ n≡d/2 (mod d) \end{matrix}} \frac{1}{n}
= \frac{1}{d} ∑_{1≤k≤x/d} \frac{1}{k} + \frac{1}{d} ∑_{0≤k≤x/d−1/2} \frac{1}{k + 1/2}
= \frac{1}{d} (log \frac{x}{d} + γ + O (\frac{d}{x})) + \frac{2}{d}
= \frac{1}{d} (log \frac{x}{d} + γ) + O (\frac{1}{x})

since the assumption d > \frac{2x}{3} implies \frac{2}{d} < \frac{3}{x}. On the other hand, since 1 ≤ \frac{x}{d} < \frac{3}{2}, we have

|\frac{2}{d} log \frac{2x}{d} + \frac{γ}{d} − \frac{1}{d} log \frac{x}{d}| ≤ \frac{1}{d} (log \frac{4x}{d} + γ) < \frac{log 6 + γ}{d} < \frac{3 (log 6 + γ)}{2x}

and thus in this case we also get

S(x, d) = \frac{2}{d} (log \frac{2x}{d} + γ) + O (\frac{1}{x}) .

If 1 ≤ d ≤ \frac{2x}{3}, since

∑_{1≤k≤x/d−1/2} \frac{1}{k + 1/2} = log \frac{x}{d} + γ − 2 + log 4 + O (\frac{d}{x})

where we used the fact that d ≤ \frac{2x}{3} implies \frac{x}{d} − \frac{1}{2} ≥ \frac{2x}{3d}, then

S(x, d) = \frac{1}{d} (log \frac{x}{d} + γ + O (\frac{d}{x})) + \frac{2}{d}

+ \frac{1}{d} (log \frac{x}{d} + γ − 2 + log 4 + O (\frac{d}{x}))

= \frac{2}{d} (log \frac{2x}{d} + γ) + O (\frac{1}{x})

completing the proof.

4 Sums of reciprocals

Lemma 8. Let f satisfying the conditions (1). Then

∑_{n≤x} \frac{1}{f(n)} = log x ∑_{d=1}^{∞} \frac{g(d)}{d} + K_{f} + O (\frac{(log x)^{λ+1}}{x} )

where

K_{f} := ∑_{d=1}^{∞} \frac{g(d)}{d} (γ − log d)  (9)

and

∑_{\begin{matrix}n≤2x \\ n even \end{matrix}} \frac{1}{f(n)} = \frac{log x}{2} ∑_{d=1}^{∞} \frac{g(d)ρ(d)}{d} + L_{f} + O (\frac{(log x)^{λ+1}}{x} )

where

L_{f} := \frac{1}{2} ∑_{d=1}^{∞} \frac{g(d)ρ(d)}{d} (γ + log \frac{ρ(d)}{d}) .  (10)

and where ρ(d) is given in (8) and λ is defined in (2). Proof. The proof of the first estimate follows the classical lines. We have

∑_{n≤x} \frac{1}{f(n)} = ∑_{n≤x} \frac{(g ⋆ 1)(n)}{n} = ∑_{n≤x} \frac{1}{n} ∑_{d|n} g(d)
= ∑_{d≤x} \frac{g(d)}{d} ∑_{k≤x/d} \frac{1}{k}
= ∑_{d≤x} \frac{g(d)}{d} (log \frac{x}{d} + γ + O (\frac{d}{x}))
= (log x + γ) ∑_{d≤x} \frac{g(d)}{d} − ∑_{d≤x} \frac{g(d) log d}{d} + O (\frac{1}{x} ∑_{d≤x} |g(d)|)

and by Lemma 6 we get

∑_{n≤x} \frac{1}{f(n)} = (log x + γ) ∑_{d=1}^{∞} \frac{g(d)ρ(d)}{d} − ∑_{d=1}^{∞} \frac{g(d) log(d)}{d} + O (\frac{(log x)^{λ+1}}{x} )

as asserted.

The second estimate is similar, with the additional use of Lemma 7 which gives

∑_{\begin{matrix}n≤2x \\ n even \end{matrix}} \frac{1}{f(n)} = ∑_{n≤x} \frac{1}{f(2n)} = ∑_{n≤x} \frac{(g ⋆ 1)(2n)}{2n}
= \frac{1}{2} ∑_{n≤x} \frac{1}{n} ∑_{d|2n} g(d) = \frac{1}{2} ∑_{d≤2x} g(d) ∑_{\begin{matrix}n≤x \\ d|2n \end{matrix}} \frac{1}{n}

= \frac{1}{2} ∑_{d≤2x} g(d) (\frac{ρ(d)}{d} (log \frac{ρ(d)x}{d} + γ) + O (\frac{1}{x}))
= \frac{1}{2}(log x + γ) ∑_{d≤2x} \frac{g(d)ρ(d)}{d} + \frac{1}{2} ∑_{d≤2x} \frac{g(d)ρ(d)}{d} log \frac{ρ(d)}{d}

+ O (\frac{1}{x} ∑_{d≤2x} |g(d)|)

= \frac{1}{2}(log x + γ) ∑_{d=1}^{∞} \frac{g(d)ρ(d)}{d} − \frac{1}{2}(log x + γ) ∑_{d>2x}^{∞} \frac{g(d)ρ(d)}{d}

+ \frac{1}{2} ∑_{d=1}^{∞} \frac{g(d)ρ(d)}{d} log \frac{ρ(d)}{d} − \frac{1}{2} ∑_{d>2x} \frac{g(d)ρ(d)}{d} log \frac{ρ(d)}{d}

+ O (\frac{1}{x} ∑_{d≤2x} |g(d)|)

and by Lemma 6 we get

∑_{\begin{matrix}n≤2x \\ n even \end{matrix}} \frac{1}{f(n)} = \frac{1}{2}(log x + γ) ∑_{d=1}^{∞} \frac{g(d)ρ(d)}{d} + \frac{1}{2} ∑_{d=1}^{∞} \frac{g(d)ρ(d)}{d} log \frac{ρ(d)}{d} + O (\frac{(log x)^{λ+1}}{x} )

completing the proof.

5 Proof of Theorem 1

5.1 First step: Asymptotic formula

From Lemma 8 we get

∑_{n≤x} \frac{(−1)^{n}}{f(n)} = 2 ∑_{\begin{matrix}n≤x \\ n even \end{matrix}} \frac{1}{f(n)} − ∑_{n≤x} \frac{1}{f(n)}
= log \frac{x}{2} ∑_{d=1}^{∞} \frac{g(d)ρ(d)}{d} − log x ∑_{d=1}^{∞} \frac{g(d)}{d} + 2L_{f} − K_{f}

+ O (\frac{(log x)^{λ+1}}{x} )

where K_{f} et L_{f} are given in (9) et (10), so that setting

C_{f} := 2L_{f} − K_{f} − log 2 ∑_{d=1}^{∞} \frac{g(d)}{d}
= ∑_{d=1}^{∞} \frac{g(d)}{d} ((ρ(d) − 1) (γ − log d) + ρ(d) log \frac{ρ(d)}{2})
= ∑_{\begin{matrix}d=1 \\ d even \end{matrix}}^{∞} \frac{g(d)}{d} (γ − log d) + ∑_{\begin{matrix}d=1 \\ d odd \end{matrix}}^{∞} \frac{g(d)ρ(d)}{d} log \frac{ρ(d)}{2}
= ∑_{d=1}^{∞} \frac{g(2d)}{2d} (γ − log 2d) − log 2 ∑_{\begin{matrix}d=1 \\ d odd \end{matrix}}^{∞} \frac{g(d)}{d}
= \frac{1}{2} ∑_{d=1}^{∞} \frac{g(2d)}{d} (γ − log d) − log 2 ∑_{d=1}^{∞} \frac{g(d)}{d}

we obtain

∑_{n≤x} \frac{(−1)^{n}}{f(n)} = log x ∑_{d=1}^{∞} \frac{g(d) (ρ(d) − 1)}{d} + C_{f} + O (\frac{(log x)^{λ+1}}{x} )
= \frac{log x}{2} ∑_{d=1}^{∞} \frac{g(2d)}{d} + C_{f} + O (\frac{(log x)^{λ+1}}{x} )

completing the proof.

5.2 Second step: Series expansions

The unique decomposition d = 2^{α}m with α ∈ ℤ^{+} and m ≥ 1 odd provides

∑_{d=1}^{∞} \frac{g(d)}{d} = ∑_{α=0}^{∞} \frac{g(2^{α})}{2^{α}} ∑_{\begin{matrix}m=1 \\ m odd \end{matrix}}^{∞} \frac{g(m)}{m}
= (1 + ∑_{α=1}^{∞} \frac{g(2^{α})}{2^{α}}) ∏_{p≥3} (1 + ∑_{α=1}^{∞} \frac{g(p^{α})}{p^{α}})
= \frac{1}{2} ∑_{α=0}^{∞} \frac{1}{f(2^{α})} ∏_{p≥3} (1 − \frac{1}{p}) (1 + ∑_{α=1}^{∞} \frac{1}{f(p^{α})})
= P_{f}

and

∑_{d=1}^{∞} \frac{g(2d)}{d} = ∑_{α=0}^{∞} \frac{g (2^{α+1})}{2^{α}} ∑_{\begin{matrix}m=1 \\ m odd \end{matrix}}^{∞} \frac{g(m)}{m}

= (g(2) + 2 ∑_{α=2}^{∞} \frac{g (2^{α})}{2^{α}}) ∏_{p≥3} (1 + ∑_{α=1}^{∞} \frac{g (p^{α})}{p^{α}})

= (∑_{α=1}^{∞} \frac{1}{f (2^{α})} − 1) ∏_{p≥3} (1 − \frac{1}{p}) (1 + ∑_{α=1}^{∞} \frac{1}{f (p^{α})})

= \frac{2 (S_{f}(2) − 1)}{S_{f}(2) + 1}P_{f}.

Similarly

∑_{d=1}^{∞} \frac{g(2d) log d}{d} = ∑_{α=0}^{∞} \frac{g (2^{α+1})}{2^{α}} ∑_{\begin{matrix}m=1 \\ m odd \end{matrix}}^{∞} \frac{g(m) log(2^{α}m)}{m}

= ∑_{α=0}^{∞} \frac{g (2^{α+1})}{2^{α}} (α log 2 ∑_{\begin{matrix}m=1 \\ m odd \end{matrix}}^{∞} \frac{g(m)}{m} + ∑_{\begin{matrix}m=1 \\ m odd \end{matrix}}^{∞} \frac{g(m) log m}{m})

= log 2 ∑_{α=0}^{∞} \frac{αg (2^{α+1})}{2^{α}} ∑_{\begin{matrix}m=1 \\ m odd \end{matrix}}^{∞} \frac{g(m)}{m} + ∑_{α=0}^{∞} \frac{g (2^{α+1})}{2^{α}} ∑_{\begin{matrix}m=1 \\ m odd \end{matrix}}^{∞} \frac{g(m) log m}{m}

= log 2 (∑_{α=1}^{∞} \frac{α − 2}{f (2^{α})}) ∏_{p≥3} (1 − \frac{1}{p}) (1 + ∑_{α=1}^{∞} \frac{1}{f (p^{α})})

+ (∑_{α=1}^{∞} \frac{1}{f (2^{α})} − 1) ∑_{\begin{matrix}m=1 \\ m odd \end{matrix}}^{∞} \frac{g(m) log m}{m}

= \frac{2 log 2 (s_{f}(2) − 2S_{f}(2))}{S_{f}(2) + 1}P_{f} + (S_{f}(2) − 1) ∑_{\begin{matrix}m=1 \\ m odd \end{matrix}}^{∞} \frac{g(m) log m}{m}.

Now since for s ∈ ℂ such that Re s > δ > 0, we have

∑_{\begin{matrix}m=1 \\ m odd \end{matrix}}^{∞} \frac{g(m)}{m^{s}} = ∏_{p≥3} (1 − \frac{1}{p^{s}} ) (1 + ∑_{α=1}^{∞} \frac{1}{p^{α(s−1)}f(p^{α})}) := G(s)

we infer

∑_{\begin{matrix}m=1 \\ m odd \end{matrix}}^{∞} \frac{g(m) log m}{m} = −G′(1).

Using the logarithmic derivative, we get

\frac{G′_{f}(s)}{G(s)} = ∑_{p≥3} log p (\frac{1}{p^{s} − 1} − \frac{∑{∞ \atop α=1} \frac{α}{p^{α(s−1)}f(p^{α})}}{1 + ∑_{α=1}^{∞} \frac{1}{p^{α(s−1)}f(p^{α})}} )

and therefore

∑_{\begin{matrix}m=1 \\ m odd \end{matrix}}^{∞} \frac{g(m) log m}{m} = −\frac{2P_{f}}{S_{f}(2) + 1} ∑_{p≥3} log p (\frac{1}{p − 1} − \frac{s_{f}(p)}{1 + S_{f}(p)})

thus completing the proof of Theorem 1.

6 Acknowledgments

We express our gratitude to the referee for his careful reading of the manuscript and the many valuable suggestions and corrections he made.

References

[1] J. Kaczorowski and K. Wiertelak, On the sum of the twisted Euler function, Internat. J. Number Theory 8 (2012), 1741–1761.

[2] E. Landau, Über die Zahlentheoretische Function φ(n) und ihre Beziehung zum Goldbachschen Satz, Nachrichten der Koniglichten Gesellschaft der Wissenschaften zu Göttingen, Mathematisch-Physikalische Klasse, 1900, 177–186.

[3] K. Wooldridge, Mean value theorems for arithmetic functions similar to Euler’s phifunction, Proc. Amer. Math. Soc. 58 (1976), 73–78.

2010 Mathematics Subject Classification: Primary 11A25; Secondary 11N37. Keywords: alternating sum, average order over a polynomial sequence.

(Concerned with sequences \underline{A001615}, \underline{A058026}, \underline{A065463}, \underline{A082695}, \underline{A211117}, and \underline{A211178}.)

Received February 9 2013; revised versions received March 3 2013; June 12 2013. Published in Journal of Integer Sequences, June 16 2013.

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