A Comment on the Sums $\sum_{n \in \mathbb{Z}} \frac{(-1)^{nk}}{(an+1)^k}$

TL;DR

This paper extends a method involving multiple integrals to evaluate a class of alternating series related to Hurwitz Zeta functions, providing explicit probabilistic and combinatorial representations.

Contribution

It generalizes Euler's integral approach to evaluate complex Hurwitz Zeta series using multidimensional integrals and geometric transformations.

Findings

Derived explicit integral representations for series $S(k,a)$.

Mapped integrals to hyperbolic polytopes via algebraic change of variables.

Provided probabilistic and combinatorial interpretations of the series.

Abstract

We recall a proof of Euler's identity $\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$ involving the evaluation of a double integral. We extend the method to find Hurwitz Zeta series of the form $S(k,a)=\sum_{n \in \mathbb{Z}} \frac{(-1)^{nk}}{(an+1)^k},$ where $a \in \mathbb{N} \setminus \lbrace 1 \rbrace$ and $k \in \mathbb{N}.$ In particular, we consider a general $k$-dimensional integral over $(0,1)^k$ that equals the series representation $S(k,a).$ Then we use an algebraic change of variables that diffeomorphically maps $(0,1)^k$ to a $k$-dimensional hyperbolic polytope. We interpret the integral as a sum of two probabilities, and find explicit representations of such probabilities with combinatorial techniques.

Equations

$$S(k,a)$$

$$\sum _{n\ge 1}\frac{1}{n^2}=\frac{{\pi }^2}{6}$$

$$I_2$$

$$\frac{1}{\sqrt{x_1{x}_2}(1-x_1{x}_2)}$$

$$I_2$$

$$=\sum _{n\ge 0}{\int }_{(0,1{)}^2}(x_1{x}_2{)}^{n-1/2}d{x}_{1}d{x}_2$$

$$=\sum _{n\ge 0}\frac{1}{(n+1/2{)}^2}$$

$$=\sum _{n\ge 0}\frac{4}{(2n+1{)}^2}$$

$$=4S(2,2),$$

$$x_1=\frac{{\xi }_1^2(1+{\xi }_2^2)}{1+{\xi }_1^2},x_2=\frac{{\xi }_2^2(1+{\xi }_1^2)}{1+{\xi }_2^2},$$

$$\det \frac{\partial (x_1,x_2)}{\partial ({\xi }_1,{\xi }_2)}$$

$${\mathbb{H}}^2$$

$$I_2$$

$$={\int }_0^{\infty }\frac{4{\cot }^{-1}({\xi }_1)}{1+{\xi }_1^2}d{\xi }_1$$

$$={\int }_0^{\infty }\frac{2\pi -4{\tan }^{-1}({\xi }_1)}{1+{\xi }_1^2}d{\xi }_1$$

$$=\frac{{\pi }^2}{2}.$$

$$S(2,2)=\frac{{\pi }^2}{8}.$$

$$\sum _{n\ge 1}\frac{1}{n^2}$$

$$=\sum _{n\ge 1}\frac{1}{(2n{)}^2}+S(2,2)$$

$$=\frac{1}{4}\zeta (2)+\frac{{\pi }^2}{8},$$

$$\sum _{n\ge 1}\frac{1}{n^2}=\frac{{\pi }^2}{6}$$

$$I_{k,a}$$

$$I_{k,a}=S(k,a).$$

$$\frac{1}{a^k}\frac{(x_1\dots x_k{)}^{-1+1/a}+(x_1\dots x_k{)}^{-1/a}}{1-(-1{)}^k{x}_1\dots x_k}$$

$$I_{k,a}$$

$$=\sum _{n\ge 0}{\int }_{(0,1{)}^k}\frac{(-1{)}^{nk}}{a^k}\left((x_1\dots x_k{)}^{n-1+1/a}+(x_1\dots x_k{)}^{n-1/a}\right)d{x}_1\dots d{x}_k$$

$$=\sum _{n\ge 0}\frac{(-1{)}^{nk}}{a^k(n+1/a{)}^k}+\sum _{n\ge 0}\frac{(-1{)}^{nk}}{a^k(n-1/a+1{)}^k}$$

$$=\sum _{n\ge 0}\frac{(-1{)}^{nk}}{(an+1{)}^k}+\sum _{n\ge 0}\frac{(-1{)}^{nk}}{(an+a-1{)}^k}$$

$$=\sum _{n\ge 0}\frac{(-1{)}^{nk}}{(an+1{)}^k}+\sum _{n\ge 0}\frac{(-1{)}^{(n+1)k}}{(-an-a+1{)}^k}$$

$$=\sum _{n\ge 0}\frac{(-1{)}^{nk}}{(an+1{)}^k}+\sum _{n\le -1}\frac{(-1{)}^{nk}}{(an+1{)}^k}$$

$$=\sum _{n\in \mathbb{Z}}\frac{(-1{)}^{nk}}{(an+1{)}^k}=S(k,a),$$

$$x_i$$

$$\det \frac{\partial (x_1,\dots ,x_k)}{\partial ({\xi }_1,\dots ,{\xi }_k)}$$

$${\mathbb{H}}^k$$

$$\frac{\partial x_i}{\partial {\xi }_j}$$

$$\det \frac{\partial (x_1,\dots ,x_k)}{\partial ({\xi }_1,\dots ,{\xi }_k)}$$

$$A_{ij}$$

Full text

1 Introduction

In this article, we evaluate Hurwitz Zeta Series of the form

[TABLE]

The values of such series can be obtained through standard techniques from Fourier Analysis and complex variables. Some specific examples of $S(k,a)$ are found in [JC, BFY, VK], with the case $a=4$ being the focal point of [BCK, NDE, ZS2010, ZS2012].

In particular, we provide an alternative method using multiple integration. We generalize the double integral proof of

[TABLE]

by Zagier and Kontsevich [ZK, p. 8 - 9], which is as follows. Consider the integral

[TABLE]

We convert the integrand into a geometric series,

[TABLE]

Replacing the integrand with the geometric series representation, we find

[TABLE]

where the interchanging of sum and integral in (1.5) follows from the Monotone Convergence Theorem. On the other hand, the change of variables

[TABLE]

has Jacobian Determinant

[TABLE]

and diffeomorphically maps $(0,1)^{2}$ to the hyperbolic triangle

[TABLE]

Hence,

[TABLE]

Thus,

[TABLE]

Finally, we can write

[TABLE]

from which we may deduce

[TABLE]

using simple algebra.

We extend this proof to find arbitrary $S(k,a).$ In particular, we evaluate the integral

[TABLE]

the generalization of (1.3) in two ways. The first way will be to convert the integrand into a geometric series and show that it is equal to $S(k,a).$ On the other hand, we will use a change of variables generalizing (1.6).

2 Evaluation of $I_{k,a}$

2.1 From Integral to Series

We will evaluate

[TABLE]

the generalization of (1.3) in two ways.

Theorem 2.1.1**.**

We have

[TABLE]

Proof.

First, we convert the integrand into a geometric series as such

[TABLE]

Replacing this geometric series representation with the integrand in $I_{k,a}$ we obtain

[TABLE]

where the interchanging of sum and integral in (2.3) follows from the Monotone Convergence Theorem. ∎

2.2 From Integral to Hyperbolic Polytope

Now, we evaluate $I_{k,a}$ directly. We use the change of variables

[TABLE]

where we cyclically index mod $k,$ that is, we have $\xi_{k+1}:=\xi_{1}.$

Theorem 2.2.1**.**

The change of variables from (2.4) has Jacobian Determinant

[TABLE]

and diffeomorphically maps $(0,1)^{k}$ to the hyperbolic polytope

[TABLE]

Proof.

The case $k=1$ is trivial. The case $k=2$ recovers the change of variables in (1.6) from the introduction; one may see that the stated results in (1.7) and (1.8) corroborate the theorem.

Suppose $k>2.$ Note that

[TABLE]

These are the entries of the Jacobian matrix $\frac{\partial(x_{1},\dots,x_{k})}{\partial(\xi_{1},\dots,\xi_{k})}$ corresponding to (2.4). Using cofactor expansion along the first row, we find that

[TABLE]

where $A$ and $B$ are $(k-1)\times(k-1)$ matrices with entries

[TABLE]

It can be seen that $A$ is lower triangular, and $B$ is upper triangular. Hence their determinants are easy to calculate using cofactor expansions on the top row of $A$ and the bottom row of $B,$ respectively. The result will simplify down to the claimed Jacobian determinant.

For the second statement, it can be shown that (2.4) is a $\mathcal{C}^{1}$ bijective map from $(0,1)^{k}$ to $\mathbb{H}^{k}$ with $\det\frac{\partial(x_{1},\dots,x_{k})}{\partial(\xi_{1},\dots,\xi_{k})}\neq 0$ in $\mathbb{H}^{k}.$ The Inverse Function Theorem guarantees on any local neighborhood in $\mathbb{H}^{k},$ we will have

[TABLE]

∎

Remark**.**

When $a=2,$ if we instead were to make the substitution $x_{i}=\sqrt{\frac{\xi_{i}^{2}(1+\xi_{i+1}^{2})}{1+\xi_{i}^{2}}}$ and then $\xi_{i}=\tan(u_{i})$ will result in us obtaining

[TABLE]

which is Calabi’s trigonometric change of variables, considered in all of [BCK, NDE, ZS2010, ZS2012, DA'K, FL, VK]. Hence, we may view (2.4) as an algebraic generalization of Calabi’s change of variables.

Hence, our two theorems and the change of variables formula imply

[TABLE]

We wish to evaluate (2.5) by mimicking the combinatorial analysis used in [VK, p. 592 - 599]

2.3 Hyperbolic Polytope and Combinatorics

We write $[m]:=\{1,\dots,m\}$ for $m\in\mathbb{N}.$ Let $\Xi_{i}$ for $i\in[k]$ be independent and identically distributed with density function

[TABLE]

Similarly, let $\Theta_{i}$ for $i\in[k]$ be independent and identically distributed with density function

[TABLE]

Theorem 2.3.1**.**

For each $i\in[k],$ both $f_{\Xi_{i}}(\xi_{i})$ and $f_{\Theta_{i}}(\theta_{i})$ are valid density functions.

Proof.

We recall the cumulative distribution function for $\Xi_{i}$ and $\Theta_{i}$ are

[TABLE]

The claim is equivalent to showing $\lim_{t\rightarrow\infty}F_{X_{i}}(t)=1$ and $\lim_{t\rightarrow\infty}F_{\Theta_{i}}(t)=1.$

According to Gradshteyn and Ryzhik [GR, Section 2.142], we see

[TABLE]

where

[TABLE]

It can be shown upon plugging in $x=0,$ and converting the cosines and sines into complex exponentials, that the right hand side of (2.8) is $-\frac{\pi}{a}\csc(\pi/a).$ It also can be shown that as $x\rightarrow\infty,$ the right hand side of (2.8) approaches $0.$ Observing these facts will allow us to deduce $\lim_{t\rightarrow\infty}F_{X_{i}}(t)=1.$

The second result follows from making the substitution $\theta_{i}\mapsto 1/\theta_{i}$ in the defining integral representation presented in (2.7) and deducing $F_{\Theta_{i}}(t)=1-F_{\Xi_{i}}(1/t).$ ∎

Our main goal is to evaluate

[TABLE]

where $\Xi_{k+1}:=\Xi_{1},\Theta_{k+1}:=\Theta_{1}.$ In words, (2.9) is the sum of the probability that all $\Xi_{i}$ have cyclically consecutive products less than $1$ and the probability that all $\Theta_{i}$ have cyclically consecutive products less than $1.$ It is easy to see through (2.5) that $S(k,a)$ is precisely the product of $\left(\frac{\pi}{a}\csc\left(\frac{\pi}{a}\right)\right)^{k}$ and (2.9).

We begin with the easy case in calculating (2.9).

Theorem 2.3.2**.**

Suppose $\Xi_{i},\Theta_{i}<1$ for each $i\in[k].$ Then (2.9) is equal to

[TABLE]

Proof.

Clearly for any $i\in[k],$ we have $\Xi_{i}\Xi_{i+1},\Theta_{i}\Theta_{i+1}<1.$ Hence (2.9) is equal to

[TABLE]

from which the result immediately follows. ∎

The nontrivial case is when there exists $i\in[k]$ such that $\Xi_{i},\Theta_{i}\geq 1.$ We wish to set up an explicit integral representation of (2.9) in this case.

Theorem 2.3.3**.**

Suppose $\Xi_{1},\dots,\Xi_{k},\Theta_{1},\dots,\Theta_{k}$ satisfy the conditions as described by their respective probability terms in (2.9). Suppose further $r_{1},\dots,r_{n}\in[k]$ are distinct with $1\leq\Xi_{r_{n}}\leq\dots\leq\Xi_{r_{1}}$ and $1\leq\Theta_{r_{n}}\leq\dots\leq\Theta_{r_{1}}.$ Then for distinct $i,j\in[n],$ we have $r_{i}$ and $r_{j}$ are pairwise cyclically nonconsecutive, that is, $|r_{i}-r_{j}|\notin\{1,k-1\}.$ In addition, $n\leq\lfloor k/2\rfloor.$

Proof.

The proofs of these statements are identical to those in [VK, Theorem 3.2, 3.3]. ∎

We now use a mechanism to set up the integral corresponding to (2.9) if $r_{1},\dots,r_{n}\in[k]$ satisfy the first statement of the previous theorem. For each $j\in[n],$ define $\alpha_{j}$ to be the number of $\Xi_{z}$ (or $\Theta_{z}$) from $\{\Xi_{r_{j}\pm 1}\}$ (or $\{\Theta_{r_{j}\pm 1}\}$) with the property that $\sup(\Xi_{z})=1/\Xi_{r_{j}}$ (or $\sup(\Theta_{z})=1/\Theta_{r_{j}}$). In words, $\alpha_{j}$ counts the number of bounds of the form $0<\Xi_{z}<1/\Xi_{r_{j}}$ (or $0<\Theta_{z}<1/\Theta_{r_{j}}$) that will appear when we set up the integral for the first probability term (or second probability term) in (2.9).

Theorem 2.3.4**.**

We have

[TABLE]

where $\delta(a,b)=1$ if $a=b$ and [math] else.

Proof.

The proof is identical to that of [VK, Theorem 3.5]. ∎

Now we are ready to set up an integral representation for (2.9) if $r_{1},\dots,r_{k}\in[k]$ satisfy the first statement in Theorem 2.3.3.

Theorem 2.3.5**.**

If $r_{1},\dots,r_{k}\in[k]$ satisfy the first statement in Theorem 2.3.3, we have (2.9) is equal to the sum of the integrals

[TABLE]

where $\psi(t)$ and $\phi(t)$ are the cumulative distribution functions defined in (2.6), (2.7), respectively.

Proof.

We already know the integral bounds for $\Xi_{r_{1}},\dots,\Xi_{r_{n}}.$ We already know there are $\sum_{j=1}^{n}\alpha_{j}$ bounds of the form $0<\Xi_{z}<1/\Xi_{r_{j}}.$ This means there are $k-n-\sum_{j=1}^{n}\alpha_{j}$ bounds of the form $0<U_{t}<1.$ Explicitly, the first probability term in (2.9) is

[TABLE]

where $dV$ is the product of the differentials $d\xi_{1},\dots d\xi_{k}$ in the appropriate order as dictated by the integral bounds. It follows that (2.10) is equal to $J_{r_{1},\dots,r_{n}}$ upon evaluating the innermost integrals.

A similar argument can be used to show that the second probability in (2.9) is equal to $K_{r_{1},\dots,r_{n}}.$ ∎

Our theorems and (2.5) give the following result

[TABLE]

where $J_{r_{1},\dots,r_{n}}$ and $K_{r_{1},\dots,r_{n}}$ are defined as in the previous theorem.