1 Introduction and statement of the main results

Given a complex number \(\alpha ,\) for any \(m\in {\mathbb{N}}=\{ 1,2,\ldots \}\) the sum of positive divisors function \(\sigma _{\alpha }(m)\) is defined as the sum of the \(\alpha\)th powers of the positive divisors of m [1]. It is usually expressed as

$$\begin{aligned} \sigma _{\alpha }(m)\,{:}{=}\,\sum _{k\mid m }k^{\alpha }, \end{aligned}$$

where k|m means that \(k\in {\mathbb{N}}\) is a divisor of m. The functions so defined are called divisor functions. They play an important role in number theory, namely within the study of divisibility properties of integers and the distribution of prime numbers. Such functions appear in connection with the Riemann zeta function

$$\begin{aligned} \zeta (s)\,{:}{=}\,\sum _{k=1}^\infty \frac{1}{k^s},\quad (\text{Re }s>1). \end{aligned}$$

Indeed, given any \(\alpha \in {\mathbb{R}},\) it is plain that

$$\begin{aligned} \sigma _\alpha (m)=\sum _{k\mid m }\left( \frac{m}{k}\right) ^{\alpha }\le m^{\alpha }\sum _{k=1}^m \frac{1}{k^\alpha }. \end{aligned}$$

In particular, for \(\alpha >1\) this yields

$$\begin{aligned} G_\alpha (m)\,{:}{=}\,\frac{\sigma _{\alpha }(m)}{m^\alpha }\le \zeta (\alpha ),\quad \forall m\in {\mathbb{N}}. \end{aligned}$$

In 1913, Grönwall [2] showed in a quite elementary way that if \(\alpha >1,\) then

$$\begin{aligned} \limsup _{m\rightarrow \infty }G_\alpha (m)= \zeta (\alpha ). \end{aligned}$$
(1)

In fact, in view of the previous inequality, immediately he derives (1) after proving that

$$\begin{aligned} \lim _{m\rightarrow \infty }G_\alpha (P_m)= \zeta (\alpha ), \end{aligned}$$

where

$$\begin{aligned} P_m\,{:}{=}\,\prod _{i=1}^mp_i^m \end{aligned}$$

is the mth power of the product of the first m prime numbers \(p_1=2,p_2=3,p_3=5, \ldots ,p_m.\)

By arguing in an alternative way, in Sect. 4 we prove the following result, which still yields Grönwall’s equation (1).

Theorem 1.1

For every real number \(\alpha >1\) one has

$$\begin{aligned} \lim _{m\rightarrow \infty }G_\alpha (m!)=\lim _{m\rightarrow \infty }G_\alpha \left( [1,2,\ldots ,m]\right) = \zeta (\alpha ), \end{aligned}$$

where \([1,2,\ldots ,m]\) is the least common multiple of \(1,2,\ldots ,m.\)

The proof of this theorem is arguably more involved than Grönwall’s one. However, a novelty of our approach lies in the fact that, unlike Grönwall, we do not use the multiplicativity of \(\sigma _{\alpha },\) i.e., \(\sigma _{\alpha }(mn)=\sigma _{\alpha }(m)\sigma _{\alpha }(n)\) for any coprime m, n [1], but we exploit the properties of a sequence of \(C^\infty\) functions whose pointwise limit is \(\sigma _{\alpha }\) (see Sect. 2).

Our main tool is the following theorem, which, to the best of our knowledge, is new in the literature.

Theorem 1.2

Let \(\eta _1, \eta _2\in {\mathbb{R}}{\setminus }{\mathbb{Z}}\) or \(\eta _1, \eta _2\in {\mathbb{Z}},\) with \(\eta _1 \le \eta _2.\) For every \(f \in L^1[\eta _1, \eta _2 ]\) which is continuous in a neighborhood of the integers, we have

$$\begin{aligned} \lim _n\sqrt{n\pi } \int _{ \eta _1 }^{ \eta _2 } \cos ^{2n}(\pi x) f(x)\text{d}x= \sum _{m=\lfloor \eta _1\rfloor }^{\lfloor \eta _2\rfloor } f(m) - \frac{f( \eta _1)\chi _{_{\mathbb{Z}}}( \eta _1)+f( \eta _2)\chi _{_{\mathbb{Z}}}( \eta _2)}{2}, \end{aligned}$$
(2)

where \(\lfloor \eta \rfloor\) denotes the integer part of \(\eta \in {\mathbb{R}}\) and \(\chi _{_{\mathbb{Z}}}\) is the characteristic function of \({\mathbb{Z}}.\)

Our hope is also that this alternative approach might provide new insights and lead to some new result in the future, such as Ingham type formulas for convolution sums involving \(\sigma _{\alpha }\) (see [3]).

1.1 Notation

For brevity, we write \(\lim _n\) instead of \(\begin{array}{c} \lim _{n\in {\mathbb{N}}} \\ {n\rightarrow \infty } \end{array},\) and the symbol \(\{\cdot \}_{n\in {\mathbb{N}}}\) for a sequence is abbreviated as \(\{\cdot \}_n.\) The power \(\left( \cos (y)\right) ^\beta\) is written as \(\cos ^\beta (y),\) and the same we do for the sine function.

Given a real number \(x\ge 1,\) in sums like \(\sum _{k=1}^{x}\) we mean that \(k\in {\mathbb{N}}\) with \(k\le \lfloor x\rfloor ,\) where \(\lfloor x\rfloor\) denotes the integer part of \(x\in {\mathbb{R}},\) i.e., the greatest integer such that \(x- \lfloor x\rfloor \ge 0.\)

The characteristic function of a set \(A\subseteq {\mathbb{R}}\) is denoted by \(\chi _{_A}.\)

2 Approximation of the divisor function

In this section a sequence of functions that approximate \(\sigma _{\alpha }\) is defined. To this end, we consider \(\sigma _{\alpha }\) for real \(\alpha >0,\) and we extend such a function to \([0, \infty )\) by letting \(\sigma _{\alpha } (x)=0\) when \(x\in [0, \infty ){\setminus }{\mathbb{N}}.\)

For any given real numbers \(\alpha > 0\) and \(M>1,\) let us consider the sequence \(\{{\mathcal{C}}_{\alpha , n}\}_n\subset C^\infty [0, M]\) defined as

$$\begin{aligned} {\mathcal{C}}_{\alpha ,n}(M;x)\,{:}{=}\, \sum _{ k=0} ^M f_n(k), \end{aligned}$$
(3)

where

$$\begin{aligned} f_n(t)=f_{\alpha ,n}(x;t)\,{:}{=}\,\left\{ \begin{array}{ll} 0 &{} \quad \text{if }t=0 \text{ or }x=0,\\ t^\alpha \cos ^{2n}\bigg (\frac{\pi x}{t} \bigg ) &{}\quad \text{otherwise}. \end{array}\right. \end{aligned}$$

The main property of the functions (3) is given by the following proposition.

Proposition 2.1

For any given real numbers \(\alpha > 0,\) \(M>1\) and \(x\in [0, M],\) the sequence \(n\rightarrow {\mathcal{C}}_{\alpha , n}(M;x)\) is non-increasing, and \(\lim _n {\mathcal{C}}_{\alpha , n}(M;x)= \sigma _\alpha (x).\)

Proof

If \(x=0,\) then this is obviously true. Let \(x\in (0,M]\) and \(k \in (0, M]\cap {\mathbb{N}}\) be fixed. If \(x\not \in {\mathbb{N}}\) or if \(x\in {\mathbb{N}}\) is not divisible by k,  then \(|\cos (\pi x/k) |<1.\) In this case, the sequence \(n\rightarrow \cos ^{2n }(\pi x/k)\) is decreasing and \(\lim _n\cos ^{2n } (\pi x/k)=0.\) On the other hand, \(\cos ^{2n }(\pi x/k)=1\) for all \(n\in {\mathbb{N}}\) if and only if \(x\in {\mathbb{N}}\) and k divides x. Consequently,

$$\begin{aligned} \lim _n {\mathcal{C}}_{\alpha , n}(M;x)= \left\{ \begin{array}{ll} \sum _{k\mid x }k^{\alpha } &{} \quad \text{if } x\in (0,M]\cap {\mathbb{N}}, \\ 0&{} \quad \text{otherwise}, \end{array}\right. \end{aligned}$$

as claimed. \(\square\)

3 Delta-like sequences and proof of Theorem 1.2

In the literature (see e.g. [4, 5]) a sequence of non-negative functions \(\{g_n\} _{n\in {\mathbb{N}}}\subset L^1({\mathbb{R}}^m)\) is known to be a delta-like sequence if

$$\begin{aligned} \lim _{n }\int _{{\mathbb{R}}^m} g_n(x)f(x)\text{d}x=f(0) \end{aligned}$$
(4)

for all \(f\in C^\infty _0({\mathbb{R}}^m).\) If, in addition, the functions \(g_n\) are smooth, they are also called mollifiers, or approximations of the identity.

A well-known set of delta-like sequences is made of non-negative functions \(g_n\in L^1({\mathbb{R}}^m)\) satisfying the properties (a) and (b) below.

  1. (a)

    For every \(\epsilon >0,\) there exists \(\delta >0\) such that \(\int _{|x|>\delta }g_n(x)\text{d}x <\epsilon .\)

  2. (b)

    \(\int _{{\mathbb{R}}^m} g_n(x)\text{d}x=1.\)

Here and in what follows, we have let \(|x|=\sqrt{x_1^2+\cdots +x_m^2}.\)

We prove the following

Proposition 3.1

If \(\{g_n\} _{n\in {\mathbb{N}}}\subset L^1({\mathbb{R}}^m),\) where each \(g_n\) is non-negative, satisfies the previous property (a), and

(b\(^{\prime }\)) \(\lim _n \int _{{\mathbb{R}}^m} g_n(x)\text{d}x=1,\)

then (4) holds for every \(f\in L^1({\mathbb{R}}^m)\cap L^\infty ({\mathbb{R}}^m)\) which is continuous in a neighborhood of \(x=0.\)

Proof

From (b\(^{\prime }\)) it follows that for every fixed \(0<\epsilon <1\) there exists \(N>0\) such that \(1-\epsilon< \int _{{\mathbb{R}}^m} g_n(x)\text{d}x < 1+\epsilon\) whenever \(n\ge N.\) The continuity of f at \(x=0\) yields \(|f(x)- f(0)|<\epsilon\) if \(|x|<\delta\) for some \(\delta >0.\)

Thus, by using (a) we see that if \(n\ge N,\) then

$$\begin{aligned} \int _{{\mathbb{R}}^m} g_n(x)\big (f(x)- f(0)\big )\text{d}x&= \int _{|x|<\delta } g_n(x)\big (f(x)- f(0)\big )\text{d}x+\int _{|x|>\delta } g_n(x)\big (f(x) - f(0)\big )\text{d}x\\ &< \epsilon \int _{|x|<\delta } g_n(x)\text{d}x + 2\Vert f\Vert _\infty \int _{|x|>\delta } g_n(x)\text{d}x\\ &< \epsilon (1+\epsilon ) + 2\Vert f\Vert _\infty \epsilon . \end{aligned}$$

We also have that

$$\begin{aligned}\int_{{\mathbb{R}}^m} g_n(x)\big(f(x)- f(0)\big)\text{d}x&\ge-\int_{|x|<\delta}g_n(x)|f(x)-f(0)|\text{d}x-\int_{|x|>\delta}g_n(x)|f(x)-f(0)|\text{d}x\\&>-\epsilon\int_{|x|<\delta}g_n(x)\text{d}x-2\Vert f\Vert_\infty\int_{|x|>\delta}g_n(x)\text{d}x\\ &>-\epsilon(1+\epsilon)-2\Vert f\Vert_\infty\epsilon.\end{aligned}$$

Since \(\epsilon\) is arbitrary, it follows that

$$\begin{aligned} 0=\lim _{n }\int _{{\mathbb{R}}} g_n(x) ( f(x)-f(0))\text{d}x&=\lim _{n }\int _{{\mathbb{R}}} g_n(x) f(x) \text{d}x-f(0)\lim _{n }\int _{{\mathbb{R}}}g_n(x) \text{d}x\\ &= \lim _{n }\int _{{\mathbb{R}}} g_n(x) f(x) \text{d}x-f(0), \end{aligned}$$

which yields (4). \(\square\)

3.1 Proof of Theorem 1.2

We need the following

Lemma 3.2

Let \(\varphi _n(x)\,{:}{=}\, \sqrt{n\pi }\cos ^{2n}(\pi x)\) and \(\eta \in (0,1).\)

If \(f\in L^1 (-\eta , 1-\eta )\) is continuous at \(x=0,\) then

$$\begin{aligned} \lim _{n} \int _{-\eta }^{1- \eta }f(x) \varphi _n(x)\text{d}x= f(0). \end{aligned}$$
(5)

If \(f\in L^1 [0,1]\) is continuous at \(x\in \{0,1\},\) then

$$\begin{aligned} \lim _{n} \int _{0}^{1 }f(x) \varphi _n(x)\text{d}x= \frac{f(0)+f(1)}{2}. \end{aligned}$$
(6)

Proof

We first show that the function \(\varphi _n \chi _{_{(-\eta , 1-\eta )}}(x)\) satisfies the properties (a) and (b\(^{\prime }\)) in Proposition 3.1.

Without loss of generality we assume \(\eta =1/2\) and for simplicity the function \(\varphi _n \chi _{_{(-1/2, 1/2 )}}\) will still be denoted by \(\varphi _n.\)

In order to prove (a), we show that the sequence \(\{\varphi _n\}_n\) converges uniformly to zero in any compact subset of \({\mathbb{R}}{\setminus }\{0\},\) i.e., \(\lim _{n}\sup _{|x| >\delta }\varphi _n(x)=0\) for every \(\delta >0.\) Since the \(\varphi _n\) are even and decreasing when \(x>0,\) it suffices to show that \(\lim _{n}\varphi _n(\delta )=0\) for every \(0<\delta <1/2.\) By the elementary inequality \(\sin ( t) \ge 2t/\pi\) for all \(t \in [0, \,\pi /2],\) we have

$$\begin{aligned} \lim _{n}\varphi _n(\delta )&= \lim _{n}\sqrt{n\pi } \cos ^{2n}(\pi \delta )\\ &= \lim _{n} \sqrt{n\pi }\big (1-\sin ^2(\pi \delta )\big )^n \le \lim _{n} \sqrt{n\pi }(1- 4 \delta ^2)^{n} =0. \end{aligned}$$

Hence, (a) is proved.

We now prove (b\(^{\prime }\)). To this end, first let us recall the beta function [6, Formulae 8.380  2, 8.384  1]

$$\begin{aligned} \beta (a,b)= \frac{\Gamma (a)\Gamma (b )}{\Gamma (a+b)}\,{:}{=}\,2\int _{0}^{\frac{\pi }{2}} \sin ^{2a-1}(t)\cos ^{2b-1}(t)\text{d}t,\quad \text{Re }{a},\text{Re }{b}\in (0,+\infty ), \end{aligned}$$

where \(\Gamma (z)\,{:}{=}\,\int _0^\infty e^{-t}t^{z-1}\text{d}t,\) \(\text{Re }{z}>0,\) is the well-known gamma function [6, Formula 8.310  1]. Since \(\Gamma (1/2)=\sqrt{\pi }\) [6, Formula 8.338  2], we can write

$$\begin{aligned} \int _{-\frac{1}{2}}^{\frac{1}{2}} \varphi _n(x)\text{d}x&=\ 2\,\sqrt{n\pi } \int _{0}^{\frac{1}{2} } \cos ^{2n}(\pi x)\text{d}x\\ &= 2\,\sqrt{\frac{n}{\pi }} \int _{0}^{\frac{\pi }{2}} \cos ^{2n}(t)\text{d}t\\ &=\sqrt{\frac{n}{\pi }}\, \beta {\left( \frac{1}{2},\, n+\frac{1}{2}\right) }\\ &=\frac{ \Gamma \left( n+\frac{1}{2}\right) }{\Gamma (n+1)}\sqrt{n}. \end{aligned}$$

Hence, (b\(^{\prime }\)) follows after observing that by Stirling’s approximation formula we get [6, Formulae 8.339  1, 2]

$$\begin{aligned} \lim _n \frac{ \Gamma (n+\frac{1}{2})}{ \Gamma (n+1)}\sqrt{n}=\sqrt{\pi }\lim _n \frac{ (2n-1)!!}{ n!2^n}\sqrt{n}=1. \end{aligned}$$

Now, since \(f\in L^1[-\frac{1}{2},\frac{1}{2}]\) is continuous at \(x=0,\) from Proposition 3.1 we conclude that

$$\begin{aligned} \lim _n \int _{-\frac{1}{2} }^{\frac{1}{2} } \varphi _n(x) f(x)\text{d}x= \lim _n\sqrt{n\pi } \int _{-\frac{1}{2} }^{\frac{1}{2} } \cos ^{2n}(\pi x) f(x)\text{d}x=f(0). \end{aligned}$$
(7)

Thus, (5) is proved.

It remains to prove (6).

Since \(f\in L^1[0,1]\) is continuous at \(x\in \{0,1\},\) the functions \(x\rightarrow f(|x|)\) and \(x\rightarrow f(1-|x|)\) are continuous at \(x=0.\) Further, \(\varphi _n(x)= \varphi _n(1-x)\) and \(\varphi _n(x)= \varphi _n(-x).\) Thus, we can write

$$\begin{aligned} \int _{0}^{1 }f(x) \varphi _n(x)\text{d}x&= \int _{0}^{\frac{1}{2}} f(x) \varphi _n(x)\text{d}x\, + \int _ {\frac{1}{2}}^1 f(x) \varphi _n(x)\text{d}x\\ &= \int _{0}^{\frac{1}{2}} f(x) \varphi _n(x)\text{d}x\, + \int _ {0}^{\frac{1}{2}} f(1-x) \varphi _n(x)\text{d}x\\ &=\int _{-\frac{1}{2}}^{0} f(-x) \varphi _n(x)\text{d}x\, + \int _ {-\frac{1}{2}}^0 f(1+x) \varphi _n(x)\text{d}x. \end{aligned}$$

From the above chain of identities it follows that

$$\begin{aligned} 2\int _{0}^{1 }f(x) \varphi _n(x)\text{d}x= \int _{-\frac{1}{2}}^{\frac{1}{2}} f(|x|) \varphi _n(x)\text{d}x\, + \int _ {-\frac{1}{2}}^{\frac{1}{2}} f(1-|x|) \varphi _n(x)\text{d}x. \end{aligned}$$

Hence, (7) yields \(2\lim _n \int _{0}^{1 }f(x) \varphi _n(x)\text{d}x =f(0)+f(1),\) as required.

Proof of Theorem 1.2

We can assume that \([\eta _1, \eta _2]= [a-\eta , b- \eta ],\) where \(a\le b\) are integers and \(0\le \eta <1.\) This is obvious if \(\eta _1, \eta _2\in {\mathbb{Z}}.\) Therefore, let us assume that \(\eta _1, \eta _2\not \in {\mathbb{Z}},\) so that \(\eta \,{:}{=}\, \lfloor \eta _1+1 \rfloor -\eta _1\in (0,1).\) If we let \(a= \lfloor \eta _1+1 \rfloor\) and \(b=\lfloor \eta _2+2 \rfloor ,\) then \(\eta _1= a-\eta\) and \(b-\eta = \lfloor \eta _2+2 \rfloor -\eta>\eta _2+1 -\eta >\eta _2,\) i.e., \(\eta _2=b-\eta -\eta '\) for some \(\eta '>0.\) Consequently, we can extend \(f\in L^1[\eta _1, \eta _2]=L^1[a-\eta , b-\eta -\eta ']\) to \(\tilde{f}\in L^1[a-\eta , b-\eta ]\) by letting \(\tilde{f}\equiv 0\) in \([b-\eta -\eta ', \ b-\eta ].\) Note that \(\tilde{f}\) is still continuous in a neighborhood of the integers. Further, the sum on the right-hand side of (2) does not change if f is replaced by \(\tilde{f}.\) Thus, without loss of generality, we can assume that f is defined in the interval \([a-\eta ,\ b-\eta ]\) and write

$$\begin{aligned} \lim _n\sqrt{n\pi } \int _{a-\eta }^{b-\eta } \cos ^{2n}(\pi x) f(x)\text{d}x&= \sum _{m=a}^{b-1} \lim _n\sqrt{n\pi }\int _{m-\eta }^{m+1-\eta } \cos ^{2n}(\pi x) f(x)\text{d}x \\ &=\sum _{m=a}^{b-1} \lim _n \int _{ -\eta }^{ 1-\eta } \varphi _n(x) f(m+x)\text{d}x, \end{aligned}$$
(8)

where \(\varphi _n\) has been introduced in Lemma 3.2. Indeed, by applying this lemma we have

$$\begin{aligned} \lim _n \int _{ -\eta }^{ 1-\eta } \varphi _n(x) f(m+x)=\left\{ \begin{array}{ll} f(m) &{} \quad \text{if }\eta >0,\\ \frac{f(m)+f(m+1)}{2}&{}\quad \text{if }\eta =0. \end{array}\right. \end{aligned}$$

Consequently, (8) becomes (2). \(\square\)

4 Proof of Theorem 1.1

For \(x,n\in {\mathbb{N}},\) with \(x>1,\) let us take \(M=2x\) for the approximants of \(\sigma _\alpha (x)\) defined in (3), namely

$$\begin{aligned} {\mathcal{C}}_{\alpha , n}(2x;x)=\sum _{k=0}^{ 2x} k^\alpha \cos ^{2n} \bigg (\frac{\pi x}{k} \bigg ). \end{aligned}$$

Recall that the function \(f_n(t)\,{:}{=}\, t^\alpha \cos ^{2n}(\pi x/ t)\) extends to a continuous function in \([0, \infty ),\) with \(f_n(0)=f_n(2x)=0.\) Further, it satisfies the hypotheses of Theorem 1.2 in [0, 2x]. Thus,

$$\begin{aligned} {\mathcal{C}}_{\alpha , n}(2x;x)&= \lim _N\ \sqrt{N\pi }\int _{0}^{ 2x } f_n(t) \cos ^{2N}(\pi t)\text{d}t +\frac{f_n(0)+f_n(2x)}{2}\\ &=\lim _N\ \sqrt{N\pi }\int _{0}^{ 2x } f_n(t) \cos ^{2N}(\pi t)\text{d}t. \end{aligned}$$

By the change of variables \(t\rightarrow xt\) one has

$$\begin{aligned} {\mathcal{C}}_{ \alpha , n}(2x;x) = x\lim _N \sqrt{N\pi }\, F_{n,N}(x), \end{aligned}$$

where

$$\begin{aligned} F_{n,N}(x){:}{=} \int_{0}^{2} f_n(xt)\cos^{2N}(\pi xt)\text{d}t, \end{aligned}$$

with

$$\begin{aligned}f_n(xt)=\left\{\begin{array}{ll} 0&{} \quad\text{if}\ t=0,\\ (xt)^\alpha\cos^{2n}\big(\frac{\pi}{t}\big)&{} \quad\text{otherwise.}\end{array}\right.\end{aligned}$$

Thus, Theorem 2.1 yields

$$\begin{aligned} \sigma _\alpha (x)=x\lim _n\lim _N \sqrt{N\pi }\, F_{n,N}(x) ,\quad \forall x\in {\mathbb{N}}. \end{aligned}$$
(9)

Since it is plain that \(f_n(xt)\ge 0\) for all \(t\in [0,2],\) we can take a sufficiently large integer m,  set \(m'=4(m!),\) and write

$$\begin{aligned} F_{n,N}(x)&\ge\sum_{k=1}^{m}\int_{\frac{1}{k}-\frac{1}{m'}}^{\frac{1}{k}+\frac{1}{m'}} \!\!f_n(xt)\cos^{2N}(\pi xt)\text{d}t\\ &\ge \, x^\alpha\sum_{k=1}^{m} \bigg(\frac{1}{k}-\frac{1}{m'}\bigg)^\alpha\int_{\frac{1}{k}-\frac{1}{m'}}^{\frac{1}{k}+\frac{1}{m'}}\!\cos^{2n}\bigg(\frac{\pi}{t}\bigg) \cos^{2N}(\pi xt)\text{d}t\\ &= x^{\alpha-1}\, \sum_{k=1}^{m} \bigg(\frac{1}{k}-\frac{1}{m'}\bigg)^\alpha \int_{\frac{x}{k}-\frac{x}{m'}}^{\frac{x}{k}+\frac{x}{m'}} \cos^{2n}\bigg(\frac{\pi x}{t}\bigg)\cos^{2N}(\pi t)\text{d}t, \end{aligned}$$

after the change of variable \(xt \rightarrow t.\) Now, let us take \(x=m!\) so that \(x/k=m!/k\in {\mathbb{N}}\) and \(\frac{x}{k}\pm \frac{x}{m'}=\frac{m!}{k}\pm \frac{1}{4}\not \in {\mathbb{Z}}\) for all \(k\in \{1,\ldots ,m\}.\) Thus, the previous inequality becomes

$$\begin{aligned} F_{n,N}(m!) \ge (m!)^{\alpha -1}\, \sum _{k=1}^{m} \bigg (\frac{1}{k}-\frac{1}{m'}\bigg )^\alpha \int _{\frac{m!}{k}-\frac{1}{4}}^{\frac{m!}{k}+\frac{1}{4} } \cos ^{2n}\bigg (\frac{\pi m!}{ t} \bigg ) \cos ^{2N}( \pi t)\text{d}t. \end{aligned}$$

Again, we apply Theorem 1.2 by taking \(f(t)= \cos ^{2n}(\pi m!/t)\) to see that

$$\begin{aligned} \lim _{N} \sqrt{N\pi }\int _{\frac{m!}{k}-\frac{1}{4}}^{\frac{m!}{k}+\frac{1}{4} } \cos ^{2n}\bigg (\frac{ \pi m!}{ t} \bigg ) \cos ^{2N}( \pi t)\text{d}t&=f\bigg (\frac{m!}{k}\bigg )+f\bigg (\frac{m!}{k}-1\bigg )\\ &=\cos ^{2n}(k\pi )+\cos ^{2n}\left( k\pi \frac{m!}{m!-k}\right) . \end{aligned}$$

Since for any given \(k\in \{1,\ldots ,m\}\) one has \(\cos ^{2n}(k\pi )=1\) and

$$\begin{aligned} \lim _{n}\cos ^{2n}\left( k\pi \frac{m!}{m!-k}\right) = \lim _{n}\cos ^{2n}\left( \pi \frac{k}{m!/k-1}\right) =0, \end{aligned}$$

we infer

$$\begin{aligned} \lim _{n} \lim _N \sqrt{N\pi }F_{n,N}(m!)\ge (m!)^{\alpha -1}\, \sum _{k=1}^{m} \bigg (\frac{1}{k}-\frac{1}{4(m!)}\bigg )^\alpha , \end{aligned}$$

which, together with (9), implies that

$$\begin{aligned} G_\alpha (m!)= \frac{\sigma _\alpha (m!)}{(m!)^\alpha } \ge \sum _{k=1}^{m}\frac{1}{k^\alpha }+\Delta _\alpha (m), \end{aligned}$$

where

$$\begin{aligned} \Delta _\alpha (m)\,{:}{=}\,\sum _{k=1}^{m}\left( \left( \frac{1}{k}-\frac{1}{4(m!)}\right) ^\alpha -\frac{1}{k^\alpha }\right) . \end{aligned}$$

Since by the mean value theorem we see that

$$\begin{aligned} |\Delta _\alpha (m)|\le \frac{\alpha /4}{m!} \sum _{k=1}^{m}\frac{1}{k^{\alpha -1}}\le \frac{\alpha /4}{(m-1)!}, \end{aligned}$$

we deduce

$$\begin{aligned} \lim _m G_\alpha (m!) = \zeta (\alpha ), \end{aligned}$$

because, as already mentioned in Sect. 1, if \(\alpha >1,\) then

$$\begin{aligned} G_\alpha (m)\le \zeta (\alpha ),\quad \forall m\in {\mathbb{N}}. \end{aligned}$$

It is plain that in order to prove the same limit for \(G_\alpha ([1,2,\ldots ,m])\) it suffices to take \(x=[1,2,\ldots ,m]\) and proceed in a completely analogous way.

Theorem 1.1 is completely proved.