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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.

Lopsy
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    Try interchange integral and summation? – qwr May 02 '15 at 22:21
  • 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. – Chappers May 02 '15 at 22:22
  • @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. – Lucian May 02 '15 at 22:51
  • 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. – Lopsy May 02 '15 at 22:54
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    @Lopsy: The sum cannot be evaluated in closed form, unless perhaps in terms of the q-polygamma function. – Lucian May 02 '15 at 22:59
  • @Lucian Thanks! That's all I needed to know. It's a pity, I was hoping for a great trick, but I'll accept the lack of a closed form. – Lopsy May 02 '15 at 23:02
  • @Lopsy: The “trick” for evaluating the integral consisted precisely in avoiding to express the sum in closed form ! – Lucian May 02 '15 at 23:04
  • A solution to the same problem (where starts from 1) in terms of the q-polygamma function can be found at this question https://math.stackexchange.com/questions/2811384/formula-for-sum-n-1-infty-frac1anbn – Sam Coutteau Jan 12 '19 at 13:47

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