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Is there a simple closed form representation of the following sum $$ \sum_{n=1}^{\infty} \frac{1}{n^s(n+c)^s}, $$ where $s, c$ are positive integers, and $s>1$?

Clearly, this converges since: $$ \sum_{n=1}^{\infty} \frac{1}{n^s(n+c)^s} < \sum_{n=1}^{\infty} \frac{1}{n^s} = \zeta(s). $$

Mathematica is able to numerically compute the answers when inputting some specific values of $s, c$, so I was wondering if there was an analytic way (or numerical shorthand) of determining the value of the sum. I would also greatly appreciate any resources which discuss such sums.

If such a closed form exists, I was also wondering if sums such as $\sum_{n=1}^{\infty} \frac{1}{n^s(n+c)^s(n+d)^s}$ and those with more terms in the denominator have similar closed forms.

Jeb Song
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2 Answers2

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Your summand can be expanded explicitly as a partial fraction in $n$: $$\frac 1{n^s(n+c)^s}=(-1)^s\sum_{i=1}^s\binom{2s-i-1}{s-1}c^{i-2s}\left(\frac{1}{(n+c)^i}+\frac{(-1)^i}{n^i}\right).$$ This is, for instance, a special case of an identity whose history is discussed in a series of papers by Koornwinder and Schlosser:

https://arxiv.org/abs/0712.2125

https://arxiv.org/abs/1205.6362

https://arxiv.org/abs/2304.10003

Summing over $n$, we note that $$\sum_{n=1}^\infty\left(\frac{1}{(n+c)^i}+\frac{(-1)^i}{n^i}\right)=-\sum_{n=1}^c \frac 1{n^i}+\sum_{n=1}^\infty\frac{1+(-1)^i}{n^i}.$$ The second term only appears when $i$ is even. Replacing $i$ by $2i$ in the corresponding terms gives \begin{multline*}\sum_{n=1}^\infty\frac 1{n^s(n+c)^s} =(-1)^{s+1}\sum_{i=1}^s\sum_{n=1}^c\binom{2s-i-1}{s-1}\frac{c^{i-2s}}{n^i}\\ +2(-1)^s\sum_{i=1}^{\lfloor s/2\rfloor}\binom{2s-2i-1}{s-1}c^{2i-2s}\zeta(2i).\end{multline*}

Hjalmar Rosengren
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The sum appears to be expressible as $$\frac{(-1)^{s+1}}{2}X_c(s)+(-1)^s\sum_{k=0}^{\left\lfloor\frac s2\right\rfloor}\frac{\binom{2s-2k-1}{s-1}}{c^{2s-2k}}\zeta(2k),$$ for some function $X_c(s)$. I'm not able to completely determine a nice way to express $X_c(s)$, but computing it for the first few $c$ gives: \begin{align*} X_1&=0\\ X_2&=1\\ X_3&=\frac{2}{2^s}\\ X_4&=\frac{2}{3^s}+\frac{1}{4^s}\\ X_5&=\frac{2}{4^s}+\frac{2}{6^s} \end{align*} The first few terms of $X_6$ are $\frac{137}{180}$, $\frac{8009}{648000}$, and $\frac{496373}{23328000}$. The denominators are all half a power of $360$, but as of now I'm unable to find $X_6$ much less a general way to express $X_c$.

CosmicOscillator
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