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

Question

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

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. — Jyrki Lahtonen, Oct 26 '20 at 21:30

score 2 · Accepted Answer · answered Oct 26 '20 at 21:29

2

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?

answered Oct 26 '20 at 21:29

Alex R.

1 Answers1