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I want to evaluate $\sum_{k=0}^{\infty} \frac{1}{(2k+1)^5}$ to prove an integral result, how could I start? Any tips are appreciated

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    $\begingroup$ IIRC you can evaluate the alternating series $\sum_k (-1)^kk^{-5}$, for examples with the aid of selected Fourier series. This one is unknown in closed form (but trivially can be written using the zeta function). Even when the exponent is three. Look up Apery's constant. $\endgroup$
    – Jyrki Lahtonen
    Oct 26, 2020 at 21:30
  • $\begingroup$ I think this answer by Ethan might help you. $\endgroup$
    – Invisible
    Oct 26, 2020 at 22:47
  • $\begingroup$ Maybe this. $\endgroup$
    – Invisible
    Oct 26, 2020 at 23:05

1 Answer 1

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Hints:

$$1.0369=\zeta(5):=\sum_{k=1}^\infty \frac{1}{k^5}=\sum_{k=1}^\infty\frac{1}{(2k)^5}+ \sum_{k=0}^\infty\frac{1}{(2k+1)^5}$$

Can you finish from here?

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    $\begingroup$ Result is $\frac{31}{32} \zeta(5)$ $\endgroup$
    – pendermath
    Oct 26, 2020 at 21:31

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