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Is there a trick for calculating sums like

$$ S(a,b) := \sum_{n=0}^{\infty}\frac{1}{a^{n}+b^{n}} $$

where $a$ and $b$ are constants?

I've run through my usual bag of tricks: reducing it to a series I already know, telescoping, realizing the sum as a Taylor series, plugging sample answers into RIES, and even some snazzy calculus tricks. Like, I figured out that

$$ \int_{a=1}^\infty \int_{b=1}^\infty \frac{S(a,b)}{ab} \ da\ db = \frac {\pi^2}{6}\log4$$

but I feel no closer to an answer. (EDIT: to be clear, I know how to get the integral above. I want to find the sum $S(a,b)$. The integral is just something I tried that doesn't seem to help.) And googling "series of reciprocals of sums of powers" works about as well as you'd expect.

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    $\begingroup$ Try interchange integral and summation? $\endgroup$
    – qwr
    May 2, 2015 at 22:21
  • $\begingroup$ Clearly the sum diverges if both of $a$ and $b$ are smaller than $1$. If $a=b$ the result is easy. So you can pull out one of them and just have to deal with $1/(1+x^n)$, $x>1$ Even this looks pretty grim. $\endgroup$
    – Chappers
    May 2, 2015 at 22:22
  • $\begingroup$ @qwr's hint is very good. We have $\displaystyle\int_1^\infty\int_1^\infty\frac1{a^n+b^n}\frac{da}a\frac{db}b ~=~ \int_0^1\int_0^1\frac1{a^{-n}+b^{-n}}\frac{da}a\frac{db}b ~=~ \frac{\ln4}{n^2}~.~$ Then see Basel problem. $\endgroup$
    – Lucian
    May 2, 2015 at 22:51
  • $\begingroup$ To clarify: I can evaluate the integral myself. I want to know how to do the sum. The integral trick is something I tried in the hopes that it would help me find the sum. $\endgroup$
    – Lopsy
    May 2, 2015 at 22:54
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    $\begingroup$ @Lopsy: The sum cannot be evaluated in closed form, unless perhaps in terms of the q-polygamma function. $\endgroup$
    – Lucian
    May 2, 2015 at 22:59

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