# Convergence of a series involving Riemann Zeta Function

I was wondering how to show whether this series converges or not:

$\sum_{m=0}^{\infty}(-1)^{m}\frac{\pi^{2m+2}}{(2m+2)!}\zeta(-2m-1)$

Numerically it converges after a few terms in wolfra alpha. But what test can one actually use to show that it converges?

thanks

• the zeta values at negative integers are given by $-B_n/(n+1)$ where $B_n$ are the Bernoulli numbers. The asymptotics of this numbers can be given straightforwardly by stirlings approximation. Can u take it from here? Dec 4, 2015 at 12:25
• I guess I will have to use the ratio test from there onwards, right? Dec 4, 2015 at 12:31
• i would suppose, yes Dec 4, 2015 at 12:40
• Is this reasoning correct? If $(-1)^{n+1} B_n \approx \frac{2(2n)!}{(2\pi)^2n}$ then $(-1)^{2n+1} B_{2n+1} \approx \frac{2(2n+1)!}{(2\pi)^{2n+1}}$ Dec 4, 2015 at 13:19
• $S=\ln\dfrac2\pi$ Dec 4, 2015 at 13:31

Ok so here is my working. I hope it is correct: The series is given by: $\sum_{m=0}^{\infty}(-1)^{m}\frac{\pi^{2m+2}}{(2m+2)!}\zeta(-2m-1)$
Therefore the coefficients are: $c_m=(-1)^{m}\frac{\pi^{2m+2}}{(2m+2)!}\zeta(-2m-1)$
and $c_{m-1}=(-1)^{m-1}\frac{\pi^{2m}}{(2m)!}\zeta(-2m+1)$
Applying the ratio test we can leave out the term $-1$:
$$\left|\frac{c_m}{c_{m-1}}\right|=\left|\frac{\frac{\pi^{2m+2}}{(2m+2)!}\zeta(-2m-1)}{\frac{\pi^{2m}}{(2m)!}\zeta(-2m+1)}\right|=\left|\frac{\pi^{2m+2}\zeta(-2m-1)(2m)!}{\pi^{2m}\zeta(-2m+1)(2m+2)!}\right|=\left|\frac{\pi^{2}\zeta(-2m-1)}{\zeta(-2m+1)(2m+2)(2m+1)}\right|$$ Now using the approximation of the Riemann Zeta function with Bernoulli numbers we get: $$=\left|\frac{\pi^{2}\frac{B_{2m+1}}{2m+2}}{\frac{B_{2m-1}}{2m}(2m+2)(2m+1)}\right|=\left|\frac{\pi^{2}B_{2m+1}2m}{B_{2m-1}(2m+2)^2(2m+1)}\right|$$ Not sure how to proceed from here.