Functions of Slow Increase and Integer Sequences

Rafael Jakimczuk División Matemática Universidad Nacional de Luján Buenos Aires Argentina

                   jakimczu@mail.unlu.edu.ar

Abstract We study some properties of functions that satisfy the condition f′(x) = o (\frac{f(x)}{x}), for x → ∞, i.e., lim_{x→∞} \frac{f′(x)}{\frac{f(x)}{x}} = 0. We call these “functions of slow increase”, since they satisfy the condition lim_{x→∞} \frac{f(x)}{x^{α}} = 0 for all α > 0. A typical example of a function of slow increase is the function f(x) = log x. As an application, we obtain some general results on sequence A_{n} of positive integers that satisfy the asymptotic formula A_{n} ∼ n^{s}f(n), where f(x) is a function of slow increase.

1 Functions of Slow Increase

Definition 1. Let f(x) be a function defined on the interval [a, ∞) such that f(x) > 0, lim_{x→∞} f(x) = ∞ and with continuous derivative f′(x) > 0. The function f(x) is of slow increase if the following condition holds.

lim_{x→∞} \frac{f′(x)}{\frac{f(x)}{x}} = 0.  (1)

Typical functions of slow increase are f(x) = log x, f(x) = log^{2} x, f(x) = log log x, f(x) = \frac{log x}{log log x} and Ψ : (0, ∞) → (0, ∞), Ψ(x) = \frac{Γ′(x)}{Γ(x)}, where Γ(x) = ∫_{0}^{∞} t^{x−1}e^{−t} dt, which generalize the harmonic sum H_{n} : N^{∗} → R, H_{n} = 1 + \frac{1}{2} + \frac{1}{3} + ⋯ + \frac{1}{n} to (0, ∞), namely H_{n} = Ψ(n + 1) + γ, where γ is Euler’s constant.

We have the following theorems.

Theorem 2. If f(x) and g(x) are functions of slow increase and C and α are positive constants then the following functions are of slow increase.

f(x) + C,  f(x) − C,  Cf(x),  f(x)g(x),  f(x)^{α},
f(g(x)),  log f(x),  f(x^{α}),  f(x^{α}g(x)),  f(x) + g(x).

If f(x) and g(x) are functions of slow increase, lim_{x→∞} \frac{f(x)}{g(x)} = ∞ and \frac{d}{dx} (\frac{f(x)}{g(x)}) > 0 then

\frac{f(x)}{g(x)} is a function of slow increase.

If h(x) is a function such that h(x) > 0, lim_{x→∞} h(x) = ∞ and with continuous derivative h′(x) > 0, then h(log x) is a function of slow increase if and only if lim_{x→∞} \frac{h′(x)}{h(x)} = 0.

If h(x) is a function such that h(x) > 0, lim_{x→∞} h(x) = ∞ and with continuous derivative h′(x) > 0, then e^{h(x)} is a function of slow increase if and only if lim_{x→∞} xh′(x) = 0.

If f(x) is a function of slow increase the following limit holds.

lim_{x→∞} \frac{log f(x)}{log x} = 0.  (2)

Proof. Use Definition 1.

Theorem 3. The function f(x) is of slow increase if and only if \frac{f(x)}{x^{α}} has negative derivative (from a certain x_{α}) for all α > 0.

Proof. We have

\frac{d}{dx} (\frac{f(x)}{x^{α}}) = \frac{f(x)}{x^{α+1}} (\frac{xf′(x)}{f(x)} − α) .  (3)

Therefore if limit (1) holds we obtain that for all α > 0,

\frac{d}{dx} (\frac{f(x)}{x^{α}}) < 0  (4)

for x > x_{α}. On the other hand, if (4) holds (for x > x_{α}), (3) gives

0 < \frac{xf′(x)}{f(x)} < α.

Consequently we obtain (note that α is arbitrary)

lim_{x→∞} \frac{xf′(x)}{f(x)} = 0.

That is, equation (1).

The following theorem justifies the term “slow increase”.

Theorem 4. If the function f(x) is of slow increase then

lim_{x→∞} \frac{f(x)}{x^{β}} = 0  (5)

for all β > 0.

Proof. Let α > 0 be such that α < β. Then \frac{f(x)}{x^{α}} has a negative derivative (for x > x_{α}), then it is decreasing, therefore it is bounded 0 < \frac{f(x)}{x^{α}} < M. So

lim_{x→∞} \frac{f(x)}{x^{β}} = lim_{x→∞} \frac{f(x)}{x^{α}}.\frac{1}{x^{β−α}} = 0.

Corollary 5. If the function f(x) is of slow increase then the following limits hold.

lim_{x→∞} \frac{f(x)}{x} = 0,  (6)

lim_{x→∞} f′(x) = 0.  (7)

Proof. Limit (6) is an immediate consequence of Theorem 4. Limit (7) is an immediate consequence of limit (6) and limit (1).

Theorem 6. If the function f(x) is of slow increase then

∑_{i=1}^{∞} i^{α}f(i)^{β} = ∞  (8)

for all α > −1 and for all β.

Proof. We have

∑_{i=1}^{∞} i^{α}f(i)^{β} = ∑_{i=1}^{∞} (i^{α+1}f(i)^{β}) \frac{1}{i}.  (9)

Now, it is well-known that

∑_{i=1}^{∞} \frac{1}{i} = ∞.  (10)

On the other hand, we have (note that α + 1 > 0)

lim_{i→∞} i^{α+1}f(i)^{β} = ∞.  (11)

Limit (11) is clearly true if β ≥ 0. If β < 0 limit (11) is a direct consequence of Theorem 2 (f(x)^{−β} is of slow increase) and Theorem 4.

Finally, equations (9), (10) and (11) give equation (8).

Theorem 7. If the function f(x) is of slow increase then the following limit holds

lim_{x→∞} \frac{∫_{a}^{x} t^{α}f(t)^{β} dt}{\frac{x^{α+1}}{α+1} f(x)^{β}} = 1  (12)

for all α > −1 and for all β.

Proof. We have (see (11))

lim_{x→∞} \frac{x^{α+1}}{α + 1}f(x)^{β} = ∞.

On the other hand, the function t^{α}f(t)^{β} is either increasing or decreasing.

Use Theorem 2 and Theorem 3 in the case α < 0, β > 0 and α > 0, β < 0. The others cases are trivial.

Consequently (8) implies,

lim_{x→∞} ∫_{a}^{x} t^{α}f(t)^{β} dt = ∞.

Now, limit (12) is a direct consequence of the L’Hospital’s rule and limit (1).

Some particular cases of this theorem are the following:

If α = 0 we have

∫_{a}^{x} f(t)^{β} dt ∼ xf(x)^{β}.  (13)

If α = 0 and β = 1 we have

∫_{a}^{x} f(t) dt ∼ xf(x).  (14)

If α = 0 and β = −1 we have

∫_{a}^{x} \frac{1}{f(t)} dt ∼ \frac{x}{f(x)}.  (15)

Theorem 8. If the function f(x) is of slow increase and C is a constant then the following limit holds

lim_{x→∞} \frac{f(x + C)}{f(x)} = 1.  (16)

Proof. If C > 0, applying the Lagrange’s Theorem we obtain

0 ≤ \frac{f(x + C) − f(x)}{f(x)} = \frac{Cf′(ξ)}{f(x)},  (x < ξ < x + C).  (17)

Equations (17) and (7) give (16). In the same way can be proved the case C < 0.

Theorem 9. If the function f(x) is of slow increase, f′(x) is decreasing and C > 0 then the following limit holds

lim_{x→∞} \frac{f(Cx)}{f(x)} = 1.  (18)

Proof. Suppose that C > 1. Applying Lagrange’s theorem we obtain

0 ≤ \frac{f(Cx) − f(x)}{f(x)} = \frac{(Cx − x)f′(ξ)}{f(x)} ≤ (C − 1)\frac{xf′(x)}{f(x)}, (x < ξ < Cx).  (19)

Equations (19) and (1) give (18).

Suppose that C < 1. Applying Lagrange’s theorem we obtain

0 ≤ \frac{f(x) − f(Cx)}{f(Cx)} = \frac{(x − Cx)f′(ξ)}{f(Cx)} ≤ \frac{1 − C}{C} \frac{Cxf′(Cx)}{f(Cx)}, (Cx < ξ < x).  (20)

Equations (20) and (1) give (18).

Theorem 10. If the function f(x) is of slow increase, f′(x) is decreasing and 0 < C_{1} ≤ g(x) ≤ C_{2} then the following limit holds.

lim_{x→∞} \frac{f(g(x)x)}{f(x)} = 1.  (21)

Proof. We have

\frac{f(C_{1}x)}{f(x)} ≤ \frac{f(g(x)x)}{f(x)} ≤ \frac{f(C_{2}x)}{f(x)}.  (22)

Equation (22) and Theorem 9 give (21).

2 Applications to Integer Sequences

In this section we consider only functions of slow increase that have decreasing derivative. Let A_{n} be a strictly increasing sequence of positive integers such that

A_{n} ∼ n^{s}f(n),  (A_{1} > 1)  (23)

and f(x) is a function of slow increase.

Let ψ(x) be the number of A_{n} that do not exceed x.

Example 11. If A_{n} = p_{n} is the sequence of prime numbers we have (Prime Number Theorem) s = 1 and f(x) = log x. If A_{n} = c_{n,k} is the sequence of numbers with k prime factors we have s = 1 and f(x) = \frac{(k−1)! log x}{(log log x)^{k−1}} (see [2]). If A_{n} = p_{n}^{2} we have s = 2 and f(x) = log^{2} x.

Remark 12. Note that: (i) Theorem 4 implies that s ≥ 1 in equation (23).

(ii) There exists a strictly increasing sequence A_{n} that satisfies (23), for example A_{n} = ⌊n^{s}f(n)⌋.

(iii) If the function g(x) is of slow increase then \frac{g(A_{n})}{g(n)} → l ⇔ \frac{g(n^{s}f(n))}{g(n)} → l and \frac{g(A_{n})}{g(n)} →

∞ ⇔ \frac{g(n^{s}f(n))}{g(n)} → ∞, because (Theorem 10) g(A_{n}) ∼ g(n^{s}f(n)).

Theorem 13. If A_{n} satisfies (23) and g(x) is a function of slow increase then the following equations hold

A_{n+1} ∼ A_{n},  (24)

lim_{n→∞} \frac{A_{n+1} − A_{n}}{A_{n}} = 0,  (25)

log A_{n+1} ∼ log A_{n},  (26)
g(A_{n+1}) ∼ g(A_{n}),  (27)
log A_{n} ∼ s log n,  (28)
log log A_{n} ∼ log log n,  (29)

lim_{x→∞} \frac{ψ(x)}{x} = 0.

Proof. Equation (24) is an immediate consequence of equation (23) and Theorem 8. Equation (25) is an immediate consequence of equation (24). Equations (26) and (27) are an immediate consequence of equation (24) and Theorem 10. Equation (28) is an direct consequence of equations (23) and (2). Equation (29) is an direct consequence of (28). The last limit is an immediate consequence of (23) ((A_{n}/n) → ∞) and (24).

Theorem 14. If A_{n} satisfies (23) and g(x) is a function of slow increase then the following equation holds (note that l ≥ 1).

g(A_{n}) ∼ lg(n) ⇔ g(ψ(x)) ∼ \frac{1}{l}g(x).  (30)

In particular (see (28) and (29))

log A_{n} ∼ s log n ⇔ log ψ(x) ∼ \frac{1}{s} log x,  (31)

log log A_{n} ∼ log log n ⇔ log log ψ(x) ∼ log log x.  (32)

Proof. We have

g(ψ(x)) ∼ \frac{1}{l}g(x) ⇒ g(ψ(A_{n})) ∼ \frac{1}{l}g(A_{n}) ⇒ g(n) ∼ \frac{1}{l}g(A_{n})
⇒ g(A_{n}) ∼ lg(n).

On the other hand

g(A_{n}) ∼ lg(n) ⇒ g(A_{n}) ∼ lg(ψ(A_{n})) ⇒ g(ψ(A_{n})) ∼ \frac{1}{l}g(A_{n}).  (33)

If A_{n} ≤ x < A_{n+1} we have

\frac{g(ψ(A_{n}))}{\frac{1}{l}g(A_{n+1})} ≤ \frac{g(ψ(x))}{\frac{1}{l}g(x)} ≤ \frac{g(ψ(A_{n}))}{\frac{1}{l}g(A_{n})}.

Now, both sides have limit 1 (see (33) and (27)).

We shall need the following well-known lemma (see [5, p. 332]). Lemma 15. Let ∑_{i=1}^{∞} a_{i} and ∑_{i=1}^{∞} b_{i} be two series of positive terms such that lim_{i→∞} \frac{a_{i}}{b_{i}} = 1. Then if ∑_{i=1}^{∞} b_{i} is divergent, the following limit holds.

lim_{n→∞} \frac{∑_{i=1}^{n} a_{i}}{∑_{i=1}^{n} b_{i}} = 1.

In the following theorem we shall obtain information on ψ(x) when s = 1 (see (23)) and f(A_{n}) ∼ f(n). Theorem 16. If f(A_{n}) ∼ f(n) then

A_{n} ∼ nf(n) ⇔ ψ(x) ∼ \frac{x}{f(x)} ⇔ ψ(x) ∼ ∫_{a}^{x} \frac{1}{f(t)} dt ⇔ ∑_{A_{i}≤x} f(A_{i}) ∼ x.  (34)

Besides if g(x) is a function of slow increase and g(A_{n}) ∼ l′g(n) then

ψ(x) ∼ \frac{∑_{A_{i}≤x} g(A_{i})^{β}}{g(x)^{β}}  (35)

for all β. Proof. We have (note that \frac{x}{f(x)} → ∞, see (6))

ψ(x) ∼ \frac{x}{f(x)} ⇒ ψ(A_{n}) ∼ \frac{A_{n}}{f(A_{n})} ⇒ n ∼ \frac{A_{n}}{f(A_{n})} ⇒ A_{n} ∼ nf(A_{n})
⇒ A_{n} ∼ nf(n).

On the other hand

A_{n} ∼ nf(n) ⇒ A_{n} ∼ ψ(A_{n})f(n) ⇒ ψ(A_{n}) ∼ \frac{A_{n}}{f(n)}
⇒ ψ(A_{n}) ∼ \frac{A_{n}}{f(A_{n})}.  (36)

If A_{n} ≤ x < A_{n+1} we have (note that \frac{x}{f(x)} is increasing, see Theorem 3)

\frac{ψ(A_{n})}{\frac{A_{n+1}}{f(A_{n+1})}} ≤ \frac{ψ(x)}{\frac{x}{f(x)}} ≤ \frac{ψ(A_{n})}{\frac{A_{n}}{f(A_{n})}}.  (37)

Now, both sides have limit 1 (see (36), (24) and (27)). Consequently

A_{n} ∼ nf(n) ⇒ ψ(x) ∼ \frac{x}{f(x)}.

On the other hand (see (15))

ψ(x) ∼ \frac{x}{f(x)} ⇔ ψ(x) ∼ ∫_{a}^{x} \frac{1}{f(t)} dt.

Note that (see (13))

∫_{a}^{n} g(x)^{β} dx ∼ ng(n)^{β}.

Therefore as g(x)^{β} is either increasing or decreasing,

∑_{i=1}^{n} g(i)^{β} = ∫_{a}^{n} g(x)^{β} dx + h(n) ∼ ng(n)^{β}.  (38)

Equation (38), g(A_{n})^{β} ∼ l^{′β}g(n)^{β} and Lemma 15 give

∑_{i=1}^{n} g(A_{i})^{β} ∼ nl^{′β}g(n)^{β}.

That is

∑_{A_{i}≤A_{n}} g(A_{i})^{β} ∼ ψ(A_{n})g(A_{n})^{β}.

Consequently

ψ(A_{n}) ∼ \frac{∑_{A_{i}≤A_{n}} g(A_{i})^{β}}{g(A_{n})^{β}} .  (39)

If A_{n} ≤ x < A_{n+1} we have (β > 0)

\frac{ψ(A_{n})}{\frac{∑_{A_{i}≤An} g(A_{i})^{β}}{g(A_{n})^{β}}} ≤ \frac{ψ(x)}{\frac{∑_{A_{i}≤x} g(A_{i})^{β}}{g(x)^{β}}} ≤ \frac{ψ(A_{n})}{\frac{∑_{A_{i}≤An} g(A_{i})^{β}}{g(A_{n+1})^{β}}} .

Now, both sides have limit 1 (see (39) and (27)). Therefore

ψ(x) ∼ \frac{∑_{A_{i}≤x} g(A_{i})^{β}}{g(x)^{β}} .

That is, equation (35). If β < 0, the proof of (35) is the same.

Consequently if g(x) = f(x) and β = 1 we find that

ψ(x) ∼ \frac{x}{f(x)} ⇔ ∑_{A_{i}≤x} f(A_{i}) ∼ x.

Example 17. Let us consider the sequence p_{n} of prime numbers, in this case we have (Prime Number Theorem) p_{n} ∼ n log n and ψ(x) = π(x) ∼ x/(log x). Let us consider the sequence c_{n,k} of numbers with k prime factors, in this case we have c_{n,k} ∼ \frac{(k−1)!n log n}{(log log n)^{k−1}} (see

Example 11) and (Landau’s Theorem) (see [1, 2]) ψ(x) ∼ \frac{x(log log x)^{k−1}}{(k−1)! log x}.

In the following general theorem we obtain information on ψ(x) if f(A_{n}) ∼ lf(n).

Theorem 16 is a particular case of this Theorem.

Theorem 18. If f(A_{n}) ∼ lf(n) then

A_{n} ∼ n^{s}f(n) ⇔ ψ(x) ∼ l^{\frac{1}{s}} \frac{x^{\frac{1}{s}}}{f(x)^{\frac{1}{s}}} ⇔ ψ(x) ∼ \frac{l^{\frac{1}{s}}}{s} ∫_{a}^{x} \frac{t^{−1+ \frac{1}{s}}}{f(t)^{\frac{1}{s}}} dt
⇔ ∑_{A_{i}≤x} f(A_{i})^{\frac{1}{s}} ∼ l^{\frac{1}{s}} x^{\frac{1}{s}} .

Besides if g(x) is a function of slow increase and g(A_{n}) ∼ l′g(n) then

ψ(x) ∼ \frac{∑_{A_{i}≤x} g(A_{i})^{β}}{g(x)^{β}}  (40)

for all β.

Proof. The proof that

A_{n} ∼ n^{s}f(n) ⇔ ψ(x) ∼ l^{\frac{1}{s}} \frac{x^{\frac{1}{s}}}{f(x)^{\frac{1}{s}}}

is the same as in Theorem 16. Now, see equation (12),

∫_{a}^{x} \frac{t^{−1+ \frac{1}{s}}}{sf(t)^{\frac{1}{s}}} dt ∼ \frac{x^{\frac{1}{s}}}{f(x)^{\frac{1}{s}}} .

Therefore

ψ(x) ∼ l^{\frac{1}{s}} \frac{x^{\frac{1}{s}}}{f(x)^{\frac{1}{s}}} ⇔ ψ(x) ∼ \frac{l^{\frac{1}{s}}}{s} ∫_{a}^{x} \frac{t^{−1+ \frac{1}{s}}}{f(t)^{\frac{1}{s}}} dt.

The proof of the equation (40) is the same as in Theorem 16. If g(x) = f(x) and β = 1/s then we find that

ψ(x) ∼ l^{\frac{1}{s}} \frac{x^{\frac{1}{s}}}{f(x)^{\frac{1}{s}}} ⇔ ∑_{A_{i}≤x} f(A_{i})^{\frac{1}{s}} ∼ l^{\frac{1}{s}} x^{\frac{1}{s}} .

Example 19. Let us consider the following sequence of positive integers (see Theorem 22)

A_{n} = ∑_{i=1}^{n} p_{i}^{k} ∼ \frac{n^{k+1}}{k + 1} log^{k} n

where k is a positive integer. In this case we have s = k + 1, f(x) = \frac{log^{k} x}{k+1} and l = (k + 1)^{k}. Consequently

ψ(x) ∼ (k + 1)\frac{x^{\frac{1}{k+1}}}{(log x)^{\frac{k}{k+1}}} .

Let us consider the sequence P_{n} of the A_{n} powers. For example, if A_{n} = p_{n} is the sequence of prime numbers, P_{n} is the sequence of prime powers . Let λ(x) be the number of P_{n} that do not exceed x.

Theorem 20. If A_{n} satisfies (23) then

λ(x) ∼ ψ(x).  (41)

Proof. The A_{i} ≤ x are A_{1}, A_{2}, …, A_{ψ(x)}. Let us write

A_{i}^{α_{i}} = x, (i = 1, 2, …, ψ(x)).

Therefore

α_{i} = \frac{log x}{log A_{i}} , (i = 1, 2, …, ψ(x)).

We have the following inequalities

ψ(x) ≤ λ(x) ≤ ∑_{i=1}^{ψ(x)}[α_{i}] ≤ ∑_{i=1}^{ψ(x)} α_{i} = log x ∑_{i=1}^{ψ(x)} \frac{1}{log A_{i}} .  (42)

Equation (28) gives

\frac{1}{log A_{n}} ∼ \frac{1}{s log n}.  (43)

Note that (see (15))

∫_{2}^{x} \frac{1}{log t} dt ∼ \frac{x}{log x}.

Now,

\frac{1}{log A_{1}} + ∑_{i=2}^{ψ(x)} \frac{1}{s log i} = \frac{1}{log A_{1}} + \frac{1}{s} ∑_{i=2}^{ψ(x)} \frac{1}{log i}

= \frac{1}{s} ∫_{2}^{ψ(x)} \frac{1}{log t} dt + O(1) ∼ \frac{ψ(x)}{s log ψ(x)}.  (44)

Equations (43), (44) and Lemma 15 give

∑_{i=1}^{ψ(x)} \frac{1}{log A_{i}} ∼ \frac{ψ(x)}{s log ψ(x)}.  (45)

Equations (42) and (45) give

ψ(x) ≤ λ(x) ≤ h(x)\frac{ψ(x) log x}{s log ψ(x)},

where h(x) → 1. That is

1 ≤ \frac{λ(x)}{ψ(x)} ≤ h(x)\frac{log x}{s log ψ(x)}.  (46)

Finally, equations (31) and (46) give (41).

Corollary 21. The following limit holds.

lim_{x→∞} \frac{∑_{i=1}^{ψ(x)} (α_{i} − [α_{i}])}{ψ(x)} = 0.

That is, the mean fractional part has limit zero.

Theorem 22. If A_{n} satisfies (23) then the following asymptotic formulas hold

∑_{i=1}^{n} A_{i}^{α} ∼ \frac{n^{sα+1}f(n)^{α}}{sα + 1} ∼ \frac{n}{sα + 1} A_{n}^{α},  (α > 0),  (47)

∑_{A_{i}≤x} A_{i}^{α} ∼ \frac{ψ(x)}{sα + 1} x^{α},  (α > 0).  (48)

Proof. Let us consider the sum

1 + 2 + ⋯ + (n′ − 1) + ∑_{i=n′}^{n} (i^{s}f(i))^{α} ,  (49)

where n′ is a positive integer on interval [a, ∞). Note that (see (23))

A_{i}^{α} ∼ (i^{s}f(i))^{α} .  (50)

Note that the function x^{s}f(x) is increasing and therefore we have

∑_{i=n′}^{n} (i^{s}f(i))^{α} = ∫_{n′}^{n} x^{sα}f(x)^{α} dx + O (n^{sα}f(n)^{α}) .  (51)

On the other hand (see (12))

∫_{n′}^{n} x^{sα}f(x)^{α} dx ∼ \frac{n^{sα+1}f(n)^{α}}{sα + 1} .  (52)

Equations (49), (51) and (52) give

1 + 2 + ⋯ + (n′ − 1) + ∑_{i=n′}^{n} (i^{s}f(i))^{α} ∼ \frac{n^{sα+1}f(n)^{α}}{sα + 1} ∼ \frac{n}{sα + 1}A_{n}^{α}.  (53)

Finally, (53), (50) and Lemma 15 give (47).

If we substitute n = ψ(A_{n}) into equation (47) and proceed as in Theorem 14 and Theorem 16 then we obtain (48).

Remark 23. Equations (47) and (48) when A_{n} = p_{n} is the sequence of prime numbers were obtained by Sálat and Znám [6], more precise formulas when α is a positive integer were obtained by Jakimczuk [3]. Equations (47) and (48) when A_{n} = c_{n,k} is the sequence of numbers with k prime factors were obtained by Jakimczuk [2].

Jakimczuk [4] proved the following theorem. Theorem 24. If A_{n} satisfies (23) then the following formulas hold

∑_{i=1}^{n} log A_{i} = s n log n − s n + n log f(n) + o(n),

lim_{n→∞} \frac{\sqrt[n]{A_{1}A_{2} … A_{n}}}{A_{n}} = \frac{1}{e^{s}} .

Proof. See [4]. In that proof we supposed that

lim_{x→∞} ∫_{a}^{x} \frac{tf′(t)}{f(t)} dt = ∞.

Consequently (L’Hospital’s rule)

lim_{x→∞} \frac{∫_{a}^{x} \frac{tf′(t)}{f(t)} dt}{x} = 0.  (54)

This supposition is unnecessary since if the integral converges then (54) also holds. Definition 25. The function of slow increase f(x) is a universal function if and only if for all sequence A_{n} that satisfies (23) we have f(A_{n}) ∼ lf(n) where l depends of the sequence A_{n}. Example 26. Equation (28) implies that f(x) = log x is an universal function, in this case l = s. Equation (29) implies that f(x) = log log x is an universal function, in this case l = 1 does not depend of the sequence A_{n}. Remark 27. Note that if f(x) and g(x) are universal functions then f(x)^{α} (α > 0), Cf(x) (C > 0) and f(x)g(x) are universal functions. If f(x)/g(x) is a function of slow increase then is an universal function. Theorem 28. If f(x) is an universal function and A_{n} satisfies (23) then we have

ψ(x) ∼ \frac{∑_{A_{i}≤x} f(A_{i})^{β}}{f(x)^{β}}

for all β. Proof. The proof is the same as in Theorem 16 and Theorem 18. Example 29. Since f(x) = log x is an universal function, we have for all sequences A_{n} satisfying (23) that

ψ(x) ∼ \frac{∑_{A_{i}≤x} log^{β} A_{i}}{log^{β} x} .

In particular, if β = 1 we have

ψ(x) ∼ \frac{∑_{A_{i}≤x} log A_{i}}{log x} .

Theorem 30. There exist functions of slow increase that are not universal functions.

Proof. We shall prove that the following function of slow increase

g(x) = e^{\frac{log x}{log log x}} ,

is not an universal function. We shall prove that there exists a sequence A_{n} that satisfies (23) and

lim_{n→∞} \frac{g(A_{n})}{g(n)} = ∞.

Since A_{n} satisfies (23) we can write

A_{n} = h_{1}(n)n^{s}f(n),

where h_{1}(n) → 1. Therefore

\frac{g(A_{n})}{g(n)} = exp (\frac{log h_{1}(n) + s log n + log f(n)}{log log n + log s + log (1 + \frac{log f(n)}{s log n} + \frac{log h_{1}(n)}{s log n})} − \frac{log n}{log log n}) . (55)

If s > 1 (55) becomes (see (2))

\frac{g(A_{n})}{g(n)} = exp (h_{2}(n)\frac{s log n}{log log n} − \frac{log n}{log log n}) ,

where h_{2}(n) → 1. That is

\frac{g(A_{n})}{g(n)} = exp (h_{3}(n)\frac{(s − 1) log n}{log log n}) ,

where h_{3}(n) → 1. Consequently we have

lim_{n→∞} \frac{g(A_{n})}{g(n)} = ∞.

This proves the theorem. In particular this limit is true if f(x) = g(x).

To complete, we shall examine the case s = 1. In this case (55) becomes (note that lim_{x→0} \frac{log(1+x)}{x} = 1)

\frac{g(A_{n})}{g(n)} = exp (\frac{log h_{1}(n) + log n + log f(n)}{log log n + h_{4}(n)\frac{log f(n)}{log n} + h_{4}(n)\frac{log h_{1}(n)}{log n}} − \frac{log n}{log log n})

= exp (\frac{log n + log f(n)}{log log n + h_{4}(n)\frac{log f(n)}{log n} + h_{4}(n)\frac{log h_{1}(n)}{log n}} − \frac{log n}{log log n} + o(1))

= exp (\frac{log log n log f(n) − h_{4}(n) log f(n) − h_{4}(n) log h_{1}(n)}{(log log n)^{2} + h_{4}(n)\frac{log log n log f(n)}{log n} + h_{4}(n)\frac{log log n log h_{1}(n)}{log n}} + o(1))

= exp (h_{5}(n)\frac{log f(n)}{log log n} + o(1)) ,

where h_{4}(n) → 1 and h_{5}(n) → 1.

For example, if f(x) = g(x) then lim_{n→∞} \frac{g(A_{n})}{g(n)} = ∞. If f(x) = log^{α} x (α > 0) then

lim_{n→∞} \frac{g(A_{n})}{g(n)} = e^{α}. If f(x) = log log x then lim_{n→∞} \frac{g(A_{n})}{g(n)} = 1.

3 Acknowledgements

The author would like to thank the anonymous referees for their valuable comments and suggestions for improving the original version of this manuscript.

References

[1] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fourth Edition, 1960.

[2] R. Jakimczuk, A note on sums of powers which have a fixed number of prime factors, J. Inequal. Pure Appl. Math. 6 (2005), Article 31.

[3] R. Jakimczuk, Desigualdades y fórmulas asintóticas para sumas de potencias de primos, Bol. Soc. Mat. Mexicana (3) 11 (2005), 5–10.

[4] R. Jakimczuk, The ratio between the average factor in a product and the last factor, Mathematical Sciences: Quarterly Journal 1 (2007), 53–62.

[5] J. Rey Pastor, P. Pi Calleja, C. Trejo, Análisis Matemático, Volume I, Editorial Kapeluz, Octava Edición, 1969.

[6] T. Sálat and S. Znám, On the sums of prime powers, Acta. Fac. Rer. Nat. Univ. Com. Math. 21 (1968), 21–25.

2000 Mathematics Subject Classification: Primary 11B99; Secondary 11N45. Keywords: Functions of slow increase, integer sequences, asymptotic formulas.

Received September 14 2009; revised version received December 21 2009. Published in Journal of Integer Sequences, December 23 2009.

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