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Let's consider the following sequence $$ a_n = \begin{cases} 0, \quad n = 2k, \, k \in \mathbb{Z} \\ \frac{1}{n^4}, \quad n = 2k+1, \, k \in \mathbb{Z} \end{cases}. $$

I would like to compute the following series $$ S = \sum \limits_{n=1}^{\infty}a_n. $$

I know that $$ \zeta(4) = \frac{\pi^4}{90}, $$ and I assume that

$$ S = \frac{\pi^4}{96}, $$

however I don't know how to obtain that result. I would appreaciate any help.

Hendrra
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    So, you want $\sum_{k=0}^\infty \frac{1}{(2k+1)^4}$. If you take $\zeta(4)=\pi^4/90$ as given, then the method used here for the $\zeta(5)$ analogue should suffice: https://math.stackexchange.com/questions/3882414/evaluating-sum-k-0-infty-frac12k15?noredirect=1 – Semiclassical Jun 29 '24 at 14:34
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    Hint: $$\sum_{k=0}^\infty x_k = \sum_{k=0}^\infty x_{2k} + \sum_{k=0}^\infty x_{2k+1}$$ when the series converge. – jjagmath Jun 29 '24 at 14:41

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Note that $$\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n^4}+2\sum\limits_{n=1}^\infty \frac{1}{(2n)^4}=\zeta(4)$$ The second term on the left is of course $\frac{\zeta(4)}{8}.$ Thus one can obtain the required sum from the equality $$S=\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n^4}+\sum\limits_{n=1}^\infty \frac{1}{(2n)^4}=\frac{15}{16}\zeta(4)$$ In fact, for any complex number $s=\sigma+it$ with $\sigma>1$ one has (using exactly the same technique) $$\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}+\sum\limits_{n=1}^\infty \frac{1}{(2n)^s}=\frac{2^s-1}{2^s}\zeta(s)$$