# Finding the closed form for $\sum_{n=1}^{\infty }\frac{\zeta (4n)}{\beta^{4n-1}}$

Finding the closed-form

$$\sum_{n=1}^{\infty }\frac{\zeta (4n)}{\beta^{4n-1}}$$ for $\beta\in(1,+\infty)$.

I learned from this site many many important things but I till need more, so I need to know if there is a closed-form of this series . Thanks for any body devised me to learn from any answer.

• Integer number? Then why don't you do even the most basic substitution to replace $a+1$ to simplify the question? Not to mention $-2$ in the denominator is also completely superfluous. 0 effort from you, -1 from me. – user2345215 Jan 9 '15 at 20:03
• I slightly modified the title and the question body, I hope it is an improvement (especially about readability). – Jack D'Aurizio Jan 9 '15 at 20:22
• Expand $\zeta(4n)$ into its well-known infinite series, then reverse the order of the two summations. – Lucian Jan 9 '15 at 21:21

Since: $$\zeta(4n)=\frac{1}{(4n-1)!}\int_{0}^{+\infty}\frac{z^{4n-1}}{e^{z}-1}\,dz \tag{1}$$ and: $$\sum_{n\geq 1}\frac{w^{4n-1}}{(4n-1)!}=\frac{\sinh w-\sin w}{2}\tag{2}$$ it follows that, for any $\beta>1$, $$\sum_{n=1}^{+\infty}\frac{\zeta(4n)}{\beta^{4n-1}} = \frac{1}{2}\int_{0}^{+\infty}\frac{\sinh(w/\beta)-\sin(w/\beta)}{e^w-1}\,dw.\tag{3}$$ For special values of $\beta$ ($\beta\in\mathbb{N}$, for instance) we can compute the last integral through the residue theorem. In general, by exploiting the inverse Laplace transform, we have: $$\sum_{n=1}^{+\infty}\frac{\zeta(4n)}{\beta^{4n-1}} = \frac{1}{2}\left(\beta - \frac{\pi}{2}\cot\frac{\pi}{\beta}-\frac{\pi}{2}\coth\frac{\pi}{\beta}\right).\tag{4}$$ We can achieve the same by considering the Taylor series of $x\cot x$ and $x\coth x$ in a neighbourhood of $z=0$.
• Wouldn't it be simpler just to use $$\sum_{n=1}^\infty\left(\sum_{k=1}^\infty{1\over k^{4n}} \right){1\over\beta^{4n-1}}=\beta\sum_{k=1}^\infty\left(\sum_{n=1}^\infty{1\over(\beta k)^{4n}}\right)$$ – Barry Cipra Jan 18 '16 at 22:12