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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

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  • $\begingroup$ 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? $\endgroup$
    – tired
    Dec 4 '15 at 12:25
  • $\begingroup$ I guess I will have to use the ratio test from there onwards, right? $\endgroup$
    – onephys
    Dec 4 '15 at 12:31
  • $\begingroup$ i would suppose, yes $\endgroup$
    – tired
    Dec 4 '15 at 12:40
  • $\begingroup$ Is this reasoning correct? If $(-1)^{n+1} B_n \approx \frac{2(2n)!}{(2\pi)^2n}$ then $(-1)^{2n+1} B_{2n+1} \approx \frac{2(2n+1)!}{(2\pi)^{2n+1}}$ $\endgroup$
    – onephys
    Dec 4 '15 at 13:19
  • $\begingroup$ $S=\ln\dfrac2\pi$ $\endgroup$
    – Lucian
    Dec 4 '15 at 13:31
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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.

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