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What can be said about $\sum_{i=1}^{\infty} \frac{1}{x^i-y^i}$ (for $|x|>1$ and $|y|>1$ and $x\neq y$)?

Is there a kind of closed formula for this?

By comparing to the geometric series, this sum should be absolutely convergent since $|x^i-y^i|\geq \left| |x|^i - |y|^i\right|=\left||x|-|y|\right|\cdot \left(|x|^{i-1} +|x|^{i-2} |y|+\ldots+|x||y|^{i-2}+|y|^{i-1}\right)\geq \left||x|-|y|\right|\cdot i \cdot q^{i-1}\geq \left||x|-|y|\right| q^{i-1}$

with $q=\operatorname{min}\{|x|,|y|\}>1$.

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  • $\begingroup$ Wolfram alpha claims to solve the special case y=1: wolframalpha.com/… $\endgroup$
    – joro
    Commented Aug 4, 2023 at 10:36
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    $\begingroup$ Related: math.stackexchange.com/q/4317994/442 $\endgroup$
    – Gerald Edgar
    Commented Aug 4, 2023 at 10:39
  • $\begingroup$ @GeraldEdgar i think letting your $q^2$ here mathoverflow.net/questions/186303/… Equal the OP’s $\frac{y}{x}$ recasts their sum with theta functions (assuming your $a=1$ and your $x$ is the reciprocal of OP’s $a$) $\endgroup$
    – Sidharth Ghoshal
    Commented Aug 5, 2023 at 15:39
  • $\begingroup$ The sum in my reference is $-\infty$ to $\infty$, not $1$ to $\infty$. $\endgroup$
    – Gerald Edgar
    Commented Aug 5, 2023 at 20:42
  • $\begingroup$ Thanks for the comments! It seems to me that the formulas in mathoverflow.net/questions/186303/… don't work for a=1. Since they run from $-\infty$ to $+\infty$ they include n=0 and for a=1 the summand is not defined. $\endgroup$
    – borntomath
    Commented Aug 8, 2023 at 17:01

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