A GENERALIZATION OF VAN DER CORPUT’S INEQUALITY

FENG QI, JIAN CAO, AND DA-WEI NIU

Abstract. In this article, van der Corput’s inequality is generalized by using the well known Euler-Maclaurin sum formula and other analytic techniques.

1. Introduction

Let S_{n} = ∑_{k=1}^{n} \frac{1}{k} and a_{n} ≥ 0 for n ∈ ℕ such that 0 < ∑_{n=1}^{∞} a_{n} < ∞. The van der Corput’s inequality [32] reads that

∑_{n=1}^{∞} ( ∏_{k=1}^{n} a_{k}^{1/k})^{1/S_{n}} < e^{1+γ} ∑_{n=1}^{∞}(n + 1)a_{n},  (1)

where γ = 0.57721566 ⋯ stands for Euler-Mascheroni’s constant. The constant e^{1+γ} in (1) is the best possible.

Two refinements of (1) were given in [16, 19] respectively as

∑_{n=1}^{∞} ( ∏_{k=1}^{n} a_{k}^{1/k})^{1/S_{n}} < e^{1+γ} ∑_{n=1}^{∞} (n − \frac{ln n}{4})a_{n}  (2)

and

∑_{n=1}^{∞} ( ∏_{k=1}^{n} a_{k}^{1/k})^{1/S_{n}} < e^{1+γ} ∑_{n=1}^{∞} e^{−1/(4n)}(n − \frac{ln n}{3n})a_{n}.  (3)

A relation between Carleman’s inequality

∑_{n=1}^{∞} ( ∏_{i=1}^{n} a_{i})^{1/n} < e ∑_{n=1}^{∞} a_{n},  (4)

see [3, 5, 12] and the references therein, where a_{n} ≥ 0 for n ∈ ℕ such that 0 < ∑_{n=1}^{∞} a_{n} < ∞ and the constant e in the right hand side of inequality (4) is the best possible, and van der Corput’s inequality (1) was established in [35] and it was presented that

∑_{n=1}^{∞} ( ∏_{k=1}^{n} a_{k}^{1/k^{α}})^{1/S_{n}(α)} < e ∑_{n=1}^{∞} e^{αn^{α−1}S_{n}(α)}a_{n}  (5)

2000 Mathematics Subject Classification. 26D15.

Key words and phrases. van der Corput’s inequality, Carleman’s inequality, Euler-Maclaurin sum formula, power mean inequality.

This paper was typeset using AMS-L^{A}TEX.

for α ∈ [0, 1] and

∑_{n=1}^{∞} ( ∏_{k=1}^{n} a_{k}^{1/k^{α}})^{1/S_{n}(α)} < e^{\frac{1}{1−α}} ∑_{n=1}^{∞}[1 − \frac{1}{2(n + 1)}]^{\frac{α}{1−α}} a_{n}  (6)

for α ∈ [0, 1), where S_{n}(α) = ∑_{k=1}^{n} \frac{1}{k^{α}}.

Another extension of (1) was obtained in [37] as follows:

∑_{n=1}^{∞} ( ∏_{k=1}^{n} a_{k}^{1/(k+β)})^{1/T_{n}(β)} < e^{1+γ_{1}(β)} ∑_{n=1}^{∞} (n + \frac{1}{2} + β)a_{n},  (7)

where β ∈ (−1, ∞), T_{n}(β) = ∑_{k=1}^{n} \frac{1}{k+β} and

γ_{1}(β) = lim_{n→∞}[∑_{k=1}^{n} \frac{1}{k + β} − ln(n + β)].  (8)

Applying β = 0 in (7) leads to

∑_{n=1}^{∞} ( ∏_{k=1}^{n} a_{k}^{1/k})^{1/S_{n}} < e^{1+γ} ∑_{n=1}^{∞} (n + \frac{1}{2})a_{n},  (9)

which improved inequality (1) clearly.

In [4], the following extension and refinement of van der Corput’s inequality (1) were obtained as follows: Let a_{n} ≥ 0 for n ∈ ℕ such that 0 ≤ ∑_{n=1}^{∞} a_{n} < ∞. Then

∑_{n=1}^{∞} [ ∏_{k=1}^{n} a_{k}^{1/\sqrt{k(k+λ)}}]^{1/U_{n}(λ)}

< e^{1+(1+λ/3)γ(λ)} ∑_{n=1}^{∞}(n + 1)^{λ/3}[1 − \frac{ln(n + 1)}{4(n + 1 + λ/2)}]a_{n}, (10)

where λ ∈ [0, ∞), U_{n}(λ) = ∑_{k=1}^{n} \frac{1}{\sqrt{k(k+λ)}} and

γ(λ) = lim_{n→∞} [U_{n}(λ) − 2 ln \frac{\sqrt{n} + \sqrt{n + λ}}{1 + \sqrt{1 + λ}}].  (11)

In particular,

∑_{n=1}^{∞} ( ∏_{k=1}^{n} a_{k}^{1/k})^{1/S_{n}} < e^{1+γ} ∑_{n=1}^{∞} n(1 − \frac{ln n}{3n − 1/4})a_{n}.  (12)

It is easy to see that inequality (12) refines inequalities (1), (2), (3) and (9). In [21], inequality (12) was refined as

∑_{n=1}^{∞} ( ∏_{k=1}^{n} a_{k}^{1/k})^{1/S_{n}} < e^{1+γ} ∑_{n=1}^{∞} e^{− \frac{6(6n+1)γ−9}{(6n+1)(12n+11)}} n(1 − \frac{ln n}{2n + ln n + 11/6})a_{n}, (13)

where a_{n} ≥ 0 for n ∈ ℕ such that 0 < ∑_{n=1}^{∞} n(1 − \frac{ln n}{2n+ln n+11/6})a_{n} < ∞. In [20], inequalities (1), (2), (3), (9), (12) and (13) were further refined as

∑_{n=1}^{∞} ( ∏_{k=1}^{n} a_{k}^{1/k})^{1/S_{n}} < e^{1+γ} ∑_{n=1}^{∞} e^{− \frac{3γ}{6n+4}} n(1 − \frac{ln n}{2n + ln n + 4/3})a_{n},  (14)

where a_{n} ≥ 0 such that 0 < ∑_{n=1}^{∞} n(1 − \frac{ln n}{2n+ln n+4/3})a_{n} < ∞.

The aim of this paper is to generalize and refine van der Corput’s inequality further by using Euler-Maclaurin sum formula and other analytic techniques.

Our main results are the following two theorems.

Theorem 1. Let a_{n} ≥ 0 for n ∈ ℕ such that 0 ≤ ∑_{n=1}^{∞} a_{n} < ∞. Then

∑_{n=1}^{∞}[∏_{k=1}^{n} a_{k}^{1/(k+η)^{λ}}]^{1/S_{n}(η,λ)}a_{n}

≤ ∑_{n=1}^{∞}[e(1 − \frac{1 + [(n + η + 1)^{λ} − (n + η)^{λ}]S_{n}(η, λ)}{2(n + η + 1)^{λ}S_{n+1}(η, λ)})]^{1+T_{n,λ,η}}a_{n}, (15)

where S_{n}(η, λ) = ∑_{k=1}^{n} \frac{1}{(k+η)^{λ}} and T_{n,λ,η} = [(n + η + 1)^{λ} − (n + η)^{λ}]S_{n}(η, λ) for η ∈ (−1, ∞) and λ ∈ [0, 1].

Remark 1. If taking λ = 0, η = 0 or λ = 1 in inequality (15) respectively, then refinements of inequalities (4), (5) or (7) are deduced respectively. This means that Theorem 1 generalizes van der Corput’s inequality (1) and Carleman’s inequality (4).

Corollary 1. Let a_{n} ≥ 0 for n ∈ ℕ such that 0 ≤ ∑_{n=1}^{∞} a_{n} < ∞. If η ∈ (−1, ∞), then

∑_{n=1}^{∞}[∏_{k=1}^{n} a_{k}^{1/(k+η)}]^{1/S_{n}(η)}a_{n} ≤ ∑_{n=1}^{∞} e^{1+γ(η)}[n + η + \frac{1}{2} − \frac{3 ln(n + η + 1)}{16}]a_{n}, (16)

where γ(η) = γ(η, 1) and

γ(η, λ) = lim_{n→∞}[∑_{k=1}^{n} \frac{1}{(k + η)^{λ}} − ∫_{1}^{n} \frac{1}{(t + η)^{λ}} dt].  (17)

Theorem 2. Let a_{n} ≥ 0 for n ∈ ℕ such that 0 ≤ ∑_{n=1}^{∞} a_{n} < ∞. If η ≥ 0, then

∑_{n=1}^{∞}[∏_{k=1}^{n} a_{k}^{1/(k+η)^{λ}}]^{1/S_{n}(η,λ)}a_{n}

≤ ∑_{n=1}^{∞}[e(1 − \frac{1}{2(n + η + 1)})]^{1+λ(k+η)^{λ−1}S_{k}(η,λ)}a_{n}. (18)

In particular, for η = 0 and λ ∈ [0, 1), inequality

∑_{n=1}^{∞} ( ∏_{k=1}^{n} a_{k}^{1/k^{λ}})^{1/S_{n}(λ)} < ∑_{n=1}^{∞}[e(1 − \frac{1}{2(n + 1)})]^{\frac{1}{1−λ}} a_{n}  (19)

holds, where S_{n}(λ) = ∑_{k=1}^{n} \frac{1}{k^{λ}}.

Remark 2. Inequalities (18) and (19) improve inequalities (5) and (6), respectively.

2. Lemmas

To prove our main results, the following lemmas are necessary.

Recall [22, 28] that a function f is called completely monotonic on an interval I if f has derivatives of all orders on I and (−1)^{k}f^{(k)}(x) ≥ 0 for all k ≥ 0 on I.

Lemma 1. Let η ∈ (−1, ∞) and λ ∈ [0, ∞), the function f(x) = \frac{1}{(x+η)^{λ}} is completely monotonic in [1, ∞) and lim_{x→∞} f^{(i)}(x) = 0 for nonnegative integer i.

Proof. Consecutive computation yields

(−1)^{n}f^{(n)}(x) = \frac{(λ)_{n}}{(x + η)^{λ+n}} > 0,

where (c)_{k} = c(c + 1) ⋯ (c + k − 1) is Pochhammer symbol.

By induction, it is easy to verify that lim_{x→∞} f^{(i)}(x) = 0 for any nonnegative integer i. The proof of Lemma 1 is complete. □

Lemma 2. Let r < 0 and n ∈ ℕ. For any given nonnegative integer m ≥ 0,

\frac{n + m}{n + m + 1} ≤ [ \begin{matrix}\frac{1}{n+m} & ∑ & {n \atop i=1} & (i + m)^{r} \\ \frac{1}{n+m+1} & ∑ & {n+1 \atop i=1} & (i + m)^{r} \end{matrix} ]^{1/r} = [ \begin{matrix}\frac{1}{n+m} & ∑ & {n+m \atop i=m+1} & i^{r} \\ \frac{1}{n+m+1} & ∑ & {n+m+1 \atop i=m+1} & i^{r} \end{matrix} ]^{1/r}. (20)

Proof. For our own convenience, let V_{r}(m, n) = ∑_{i=1}^{n} (i + m)^{r} = ∑_{i=m+1}^{n+m} i^{r}. Then inequality (20) is equivalent to

\frac{V_{r}(m, n)}{V_{r}(m, n + 1)} ≤ (\frac{n + m}{n + m + 1})^{r+1}.  (21)

For r ≤ −1, inequality (21) holds obviously.

For 0 > r > −1, inequality (21) can be rewritten as

V_{r}(m, n) ≤ \frac{(n + m)^{r+1}(n + m + 1)^{r}}{(n + m + 1)^{r+1} − (n + m)^{r+1}} .  (22)

When n = 1, it is easy to check up that inequality (22) is true. Now assume that inequality (22) is valid for some positive integer n > 1. By induction, it is sufficient to show that inequality (22) validates for n + 1, that is,

V_{r}(m, n + 1) ≤ \frac{(n + m + 1)^{r+1}(n + m + 2)^{r}}{(n + m + 2)^{r+1} − (n + m + 1)^{r+1}} .  (23)

By inductive hypotheses and V_{r}(m, n + 1) = V_{r}(m, n) + (n + m + 1)^{r}, we have

\begin{aligned}V_{r}(m, n + 1) & ≤ \frac{(n + m)^{r+1}(n + m + 1)^{r}}{(n + m + 1)^{r+1} − (n + m)^{r+1}} + (n + m + 1)^{r} \\ & = \frac{(n + m + 1)^{2r+1}}{(n + m + 1)^{r+1} − (n + m)^{r+1}} . \end{aligned}  (24)

Therefore, in order to prove inequality (23), it suffices to show the right term of (23) is not less than the term in the second line of (24), that is,

\frac{(n + m + 1)^{r}}{(n + m + 2)^{r}} ≤ \frac{(n + m + 1)^{r+1} − (n + m)^{r+1}}{(n + m + 2)^{r+1} − (n + m + 1)^{r+1}} = \frac{(n + m + ξ)^{r}}{(n + m + 1 + ξ)^{r}} , (25)

where ξ ∈ (0, 1) is deduced by using Cauchy’s mean value theorem in the second term above, which sounds apparently. Thus, the proof of Lemma 2 is complete. □

Remark 3. It is remarked that some analogies of inequality (20) in Lemma 2 have been investigated extensively in [1, 2, 6, 7, 8, 9, 10, 11, 13, 14, 15, 18, 23, 24, 25, 26, 27, 29, 30, 31, 33, 34, 39] and the reference therein.

Lemma 3 ([17, 35, 36]). Euler-Maclaurin’s formula states that

∑_{k=1}^{n} f(k) = ∫_{1}^{n} f(x) dx + \frac{f(n) + f(1)}{2} + ∫_{1}^{n} ρ_{1}(x)f′(x) dx,  (26)

where ρ_{1}(x) = x − [x] + \frac{1}{2} is Bernoulli’s function and f ∈ C^{1}[1, ∞). Furthermore, if (−1)^{i}f^{(i)}(x) > 0 for x ∈ [n, ∞) and lim_{x→∞} f^{(i)}(x) = 0 for i = 1, 2, 3, then

∫_{n}^{∞} ρ_{1}(x)f′(x) dx = −\frac{1}{12}f′(n)ϵ, 0 < ϵ < 1.  (27)

Lemma 4. For n ∈ ℕ, η ∈ (−1, ∞) and λ ∈ [0, 1),

S_{n}(η, λ) < \frac{(n + η + 1)^{1−λ} − (1 + η)^{1−λ}}{1 − λ} + γ(η, λ),  (28)

where S_{n}(η, λ) and γ(η, λ) are defined in Theorem 1 and by (17) respectively. Moreover,

S_{n}(η, 1) < ln(n + η + 1) − ln(1 + η) + γ(η, 1).  (29)

Proof. It is clear that Lemma 1 allows us to apply Euler-Maclaurin’s formula (26) and formula (27) to f(x) = \frac{1}{(x+η)^{λ}}. From this, if λ ∈ (0, 1), it follows that

\begin{aligned}S_{n}(η, λ) & = \frac{(n + η + 1)^{1−λ} − (1 + η)^{1−λ}}{1 − λ} \\ & + \frac{1}{2}[\frac{1}{(n + η)^{λ}} + \frac{1}{(1 + η)^{λ}} ] + ∫_{1}^{n} ρ_{1}(x)[\frac{1}{(x + η)^{λ}} ]′ dx, \end{aligned}

∫_{n}^{∞} ρ_{1}(x)[\frac{1}{(x + η)^{λ}} ]′ dx = −\frac{1}{12}[\frac{1}{(x + η)^{λ}} ]′ϵ = \frac{λϵ}{12(n + η)^{1+λ}} ,

where 0 < ϵ < 1, and

γ(η, λ) = \frac{1}{2(1 + η)^{λ}} + ∫_{1}^{∞} ρ_{1}(x)[\frac{1}{(x + η)^{λ}} ]′ dx.

Therefore,

\begin{aligned}S_{n}(η, λ) & = \frac{(n + η)^{1−λ} − (1 + η)^{1−λ}}{1 − λ} + \frac{1}{2(n + η)^{λ}} + γ(η, λ) − \frac{λϵ}{12(n + η)^{1+λ}} \\ & < \frac{(n + η)^{1−λ} − (1 + η)^{1−λ}}{1 − λ} + \frac{1}{2(n + η)^{λ}} + γ(η, λ), \end{aligned}

and then

S_{n}(η, λ) = ∑_{k=1}^{n+1} \frac{1}{(k + η)^{λ}} − \frac{1}{(n + η + 1)^{λ}}

< \frac{(n + η + 1)^{1−λ} − (1 + η)^{1−λ}}{1 − λ} − \frac{1}{2(n + η + 1)^{λ}} + γ(η, λ)

< \frac{(n + η + 1)^{1−λ} − (1 + η)^{1−λ}}{1 − λ} + γ(η, λ),

Inequality (28) follows for λ ∈ (0, 1).

For λ = 0, inequality (28) holds apparently.

Since

lim_{λ→1} \frac{(n + η)^{1−λ} − (1 + η)^{1−λ}}{1 − λ} = ln(n + η) − ln(1 + η)

and \frac{1}{x} < 2 ln(1 + \frac{1}{x}) with x ≥ 1, then

S_{n}(η, 1) = ∑_{k=1}^{n} \frac{1}{k + η}

< ln(n + η) − ln(1 + η) + \frac{1}{2(n + η)} + γ(η, 1)

< ln(n + η + 1) − ln(1 + η) + γ(η, 1).

The proof of Lemma 2 is complete. □

Lemma 5. For x ∈ [0, ∞) and α ∈ [0, ∞),

[1 − \frac{1}{2(x + 1 + α)}]^{1+ln(x+α+1)} < 1 − \frac{1}{2(n + α + 1)} − \frac{3 ln(x + α + 1)}{16(x + 1 + α)}. (30)

Proof. By (1 − \frac{1}{t})^{−t} > e for t > 1, e^{t} < 1 + t + \frac{t^{2}}{2} for t < 0 and x + α > ln(x + α) with α ≥ 0, it follows that

[1 − \frac{1}{2(x + 1 + α)}]^{1+ln(x+α+1)} < [1 − \frac{1}{2(x + 1 + α)}](\frac{1}{e})^{\frac{ln(x+α+1)}{2(x+1+α)}}

< [1 − \frac{1}{2(x + 1 + α)}][1 − \frac{ln(x + α + 1)}{2(x + 1 + α)} + \frac{ln^{2}(x + α + 1)}{8(x + 1 + α)^{2}}]

< [1 − \frac{1}{2(x + 1 + α)}][1 − \frac{3 ln(x + α + 1)}{8(x + α + 1)}]

< 1 − \frac{1}{2(x + α + 1)} − \frac{3 ln(x + α + 1)}{16(x + α + 1)}.

The proof of Lemma 5 is complete. □

Lemma 6. For k ∈ ℕ, η ∈ [0, ∞) and λ ∈ [0, 1], we have

\begin{aligned}B_{k}(η, λ) & ≜ [\frac{(k + η + 1)^{λ} S_{k+1}(η, λ)}{(k + η)^{λ} S_{k}(λ)}]^{(k+η)^{λ} S_{k}(λ)} \\ & ≤ [e(1 − \frac{1}{2(k + 1 + η)})]^{1+λ(k+η)^{λ−1} S_{k}(η,λ)}. \end{aligned}  (31)

Proof. For k ∈ ℕ,

B_{k}(η, λ) = \{1 + \frac{1 + [(k + η + 1)^{λ} − (k + η)^{λ}]S_{k}(η, λ)}{(k + η)^{λ} S_{k}(λ)}\}^{(k+η)^{λ} S_{k}(η,λ)} ≜ C_{k}^{h(k)},

where

C_{k} = [1 + \frac{1}{g(k)}]^{g(k)},  g(k) = \frac{(k + η)^{λ} S_{k}(η, λ)}{h(k)}

A GENERALIZATION OF VAN DER CORPUT’S INEQUALITY 7

and

h(k, λ) = 1 + [(k + η + 1)^{λ} − (k + η)^{λ}]S_{k}(η, λ).

By Lemma 2, it is easy to see that

g(k, λ) + 1 = \frac{(k + η + 1)^{λ} S_{k+1}(η, λ)}{1 + [(k + η + 1)^{λ} − (k + η)^{λ}]S_{k}(η, λ)} ≤ k + η + 1 .  (32)

By

(1 + \frac{1}{x})^{x} < e[1 − \frac{1}{2(x + 1)}]  (33)

in [38] and inequality (32), it is deduced that

C_{k} = [1 + \frac{1}{g(k)}]^{g(k)} ≤ e\{1 − \frac{1}{2[g(k) + 1]}\} ≤ e[1 − \frac{1}{2(k + 1 + η)}].  (34)

For λ ∈ [0, 1], by using Bernoulli’s inequality, we have

h(k, λ) ≤ 1 + (k + η)^{λ}[(1 + \frac{1}{k + η} )^{λ} − 1]S_{k}(η, λ) ≤ 1 + λ(k + η)^{λ−1} S_{k}(η, λ). (35)

Hence, from inequalities (34) and (35), it is showed that

B_{k}(η, λ) ≤ \{e[1 − \frac{1}{2(k + 1 + η)}]\}^{h(k,λ)}

≤ \{e[1 − \frac{1}{2(k + 1 + η)}]\}^{1+λ(k+η)^{λ−1} S_{k}(η,λ)}.

The proof of Lemma 6 is complete. □

3. Proofs of theorems

Proof of Theorem 1. Setting c_{k} > 0 for 1 ≤ k ≤ n and letting

[∏_{k=1}^{n} c_{k}^{1/(k+η)^{λ}}]^{−1/S_{n}(η,λ)} = \frac{1}{(n + η + 1)^{λ}S_{n+1}(η, λ)},

then

c_{k} = \frac{[(k + η + 1)^{λ}S_{k+1}(η, λ)]^{(k+η)^{λ}S_{k}(η,λ)}}{[(k + η)^{λ}S_{k}(η, λ)]^{(k+η)^{λ}S_{k−1}(η,λ)}} .  (36)

Using the discrete weighted arithmetic-geometric mean inequality and (36) and interchanging the order of summation yields

8 F. QI, J. CAO, AND D.-W. NIU

∑_{n=1}^{∞}[∏_{k=1}^{n} a_{k}^{1/(k+η)^{λ}}]^{1/S_{n}(η,λ)}

\begin{matrix}= ∑_{n=1}^{∞}[∏_{k=1}^{n}(c_{k}a_{k})^{1/(k+η)^{λ}}]^{1/S_{n}(η,λ)}[∏_{k=1}^{n} c_{k}^{1/(k+η)^{λ}}]^{−1/S_{n}(η,λ)} \\ ≤ ∑_{n=1}^{∞} ∑_{k=1}^{n} \frac{1}{(k + η)^{λ} S_{n}(η, λ)}c_{k}a_{k} \frac{1}{(k + η + 1)^{λ} S_{n+1}(η, λ)} \\ = ∑_{k=1}^{∞} \frac{1}{(k + η)^{λ}} c_{k}a_{k} ∑_{n=k}^{∞} \frac{1}{(k + η + 1)^{λ}S_{n}(η, λ)S_{n+1}(η, λ)} \\ = ∑_{k=1}^{∞} \frac{1}{(k + η)^{λ}} c_{k}a_{k} ∑_{n=k}^{∞}[\frac{1}{S_{n}(η, λ)} − \frac{1}{S_{n+1}(η, λ)}] \\ = ∑_{k=1}^{∞} \frac{1}{(k + η)^{λ}} c_{k}a_{k} \frac{1}{S_{k}(η, λ)} \\ = ∑_{k=1}^{∞}[\frac{(k + η + 1)^{λ}S_{k+1}(η, λ)}{(k + η)^{λ}S_{k}(λ)}]^{(k+η)^{λ}S_{k}(λ)}a_{k}. \end{matrix} (37)

Applying (33) and the left side of inequality (32) in the final line of (37) gives inequality (15). The proof of Theorem 1 is complete. □ Proof of Corollary 1.

\begin{aligned}B_{k}(η) & = [\frac{(k + η + 1) S_{k+1}(η)}{(k + η) S_{k}(η)}]^{(k+η) S_{k}(η)} \\ & = \{[1 + \frac{1}{(k + η)S_{k}(η)/(S_{k}(η) + 1)}]^{(k+η)S_{k}(η)/(S_{k}(η)+1)}\}^{S_{k}(η)+1} \\ & < \{e[1 − \frac{S_{k}(η) + 1}{2(k + η + 1)S_{k}(η)}]\}^{S_{k}(η)+1} \\ & < \{e[1 − \frac{1}{2(k + η + 1)}]\}^{1+ln(k+η+1)+γ(η)} \\ & < e^{1+γ(η)}(k + η + 1)[1 − \frac{1}{2(k + η + 1)}]^{1+ln(k+η+1)} \\ & < e^{1+γ(η)}[k + η + \frac{1}{2} − \frac{3 ln(k + η + 1)}{16}] \end{aligned} (38)

Taking λ = 1 in inequality (37) yields

\begin{aligned}∑_{n=1}^{∞} ( ∏_{k=1}^{n} a_{k}^{1/(k+η)})^{1/S_{n}(η)} & ≤ ∑_{n=1}^{∞}[\frac{(k + η + 1) S_{k+1}(η)}{(k + η) S_{k}(η)}]^{(k+η) S_{k}(η)}a_{n} \\ & < ∑_{n=1}^{∞} e^{1+γ(η)}[n + η + \frac{1}{2} − \frac{3 ln(n + η + 1)}{16}]a_{n} \end{aligned} (39)

The proof of Corollary 1 is complete. □

Proof of Theorem 2. Applying Lemma 6 in (37) gives inequality (18) clearly. For λ = 0, since S_{n}(0) = n, inequality (19) is reduced to Carleman’s inequality

∑_{n=1}^{∞} ( ∏_{k=1}^{n} a_{k})^{n} < ∑_{n=1}^{∞} e[1 − \frac{1}{2(n + 1)}]a_{n}.  (40)

For λ ∈ (0, 1),

S_{n}(η, λ) = ∑_{k=1}^{n} \frac{1}{(k + η)^{λ}} < ∫_{0}^{n} \frac{1}{(t + η)^{λ}} dt = \frac{(n + η)^{1−λ}}{1 − λ} − \frac{η^{1−λ}}{1 − λ}. (41)

Taking η = 0 in (18) and (41) yields inequality (41). The proof of Theorem 2 is complete. □

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(F. Qi) Research Institute of Mathematical Inequality Theory, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China

   E-mail address: qifeng618@gmail.com, qifeng618@hotmail.com, qifeng618@msn.com, qifeng618@qq.com,
qifeng@hpu.edu.cn, fengqi618@member.ams.org
   URL: http://rgmia.vu.edu.au/qi.html

(J. Cao) School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China

   E-mail address: 21caojian@163.com, goodfriendforeve@163.com

(D.-W. Niu) School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China

   E-mail address: nnddww@163.com, nnddww@hotmail.com, nnddww@gmail.com, nnddww@tom.com