# A tough series: $\sum_{k=1}^{\infty}\frac{\zeta(2k)}{(k+1)(2k+1)}$, need help

I was doing a integral which ends up with a tough series part: $$\sum_{k=1}^{\infty}\frac{\zeta(2k)}{(k+1)(2k+1)}$$ Mathematica says $$\frac12$$ Which agrees with the anwer...Anyone know how to evaluate this?

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Isn't there an exact expression for the zeta function of even integers? – Ron Gordon Feb 15 at 12:20
@rlgordonma, I m not familiar with zeta :(. – Ryan Feb 15 at 12:21

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May first idea is partial fraction decomposition – Dominic Michaelis Feb 15 at 12:32
$$\frac{\sin(\pi x)}{\pi x}=\prod_{k=1}^\infty \left(1-\frac{x^2}{k^2} \right)$$ $$\begin{aligned}\log \left[ \frac{\sin(\pi x)}{\pi x}\right] &=\sum_{k=1}^\infty\log\left(1-\frac{x^2}{k^2}\right) \\ &= \sum_{k=1}^\infty \left( -\sum_{n=1}^\infty \frac{x^{2n}}{k^{2n}n}\right) \\ &= - \sum_{n=1}^\infty \frac{x^{2n}}{n}\sum_{k=1}^\infty \frac{1}{k^{2n}} \\ &= -\sum_{n=1}^\infty \frac{x^{2n}}{n}\zeta(2n)\end{aligned}$$ $$\displaystyle \ln(\sin \pi x)=\ln(\pi x)-\sum_{n=1}^\infty \frac{x^{2n}}{n}\zeta(2n)$$ We can try differentiating or integrating this equation to bring it into the required form. – Integral Feb 15 at 13:24
@ShobhitBhatnagar That's quite a voluminous comment. Maybe this would be better in the answer zone. – julien Feb 15 at 13:30
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## 2 Answers

Using the definition of $\zeta$ and exchanging the order of summation since everything is positive, one sees that the sum $S$ to be computed is $$S=\sum_{n=1}^{+\infty}\left(n^2u(\tfrac1n)-1\right),\quad\text{with}\quad u(t)=\sum\limits_{k=0}^{+\infty}\frac{2t^{2k+2}}{(2k+1)(2k+2)}.$$ From here, only elementary analysis is required. To compute $u$, note that $u(0)=u'(0)=0$ and $$u''(t)=\sum\limits_{k=0}^{+\infty}2t^{2k}=\frac2{1-t^2}=\frac1{1-t}+\frac1{1+t},$$ hence, for every $|t|\leqslant1$, $$u(t)=(1+t)\log(1+t)+(1-t)\log(1-t).$$ This yields, for each integer $n\geqslant1$, $$n^2u(\tfrac1n)=n(n+1)\log(n+1)+n(n-1)\log(n-1)-2n^2\log n,$$ hence, for every $N\geqslant1$, $$\sum_{n=1}^Nn^2u(\tfrac1n)=\sum_{n=1}^{N+1}n(n-1)\log(n)+\sum_{n=1}^{N-1}n(n+1)\log(n)-\sum_{n=1}^N2n^2\log n,$$ that is, $$\sum_{n=1}^Nn^2u(\tfrac1n)=N(N+1)\log(1+\tfrac1N).$$ Since $\log(1+x)=x-\frac12x^2+o(x^2)$ when $x\to0$, when $N\to\infty$, the RHS is $$N(N+1)\left(\tfrac1N-\tfrac12\tfrac1{N^2}+o\left(\tfrac1{N^2}\right)\right)=N+\tfrac12+o(1).$$ This proves (rigorously) that $$S=\tfrac12.$$

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Cool, nice to have something rigorous! – achille hui Feb 15 at 14:18

Interesting, here is my attempt. Nothing is rigorous.

Start with $$\sum_{k=1}^{\infty} \frac{x^{2k-1}}{\Gamma(2k)(k+1)(2k+1)} = \sum_{k=1}^{\infty} \frac{4 k x^{2k-1}}{(2k+2)!} = 2\frac{d}{dx}\left[\sum_{k=1}^{\infty}\frac{x^{2k}}{(2k+2)!}\right] = 2\frac{d}{dx}\left[\frac{\cosh x - 1 - \frac{x^2}{2}}{x^2}\right] = \frac{e^x - e^{-x}}{x^2} - 2 \frac{e^x + e^{-x} - 2}{x^3}$$ Divide both sides by $e^x - 1$, integrate and use the identity for $\Re(s) > 1$:

$$\zeta(s)\Gamma(s) = \int_0^{\infty}\frac{x^{s-1}}{e^x-1} dx$$

We get: $$\sum_{k=1}^{\infty} \frac{\zeta(2k)}{(k+1)(2k+1)} = \int_0^{\infty} \left[\frac{1+e^{-x}}{x^2} - 2 \frac{1-e^{-x}}{x^3}\right] dx = \int_0^{\infty} \frac{d}{dx}\left[-\frac{1}{x} + \frac{1-e^{-x}}{x^2}\right]dx\\ = -\lim_{x\to 0} \left[-\frac{1}{x} + \frac{1-e^{-x}}{x^2}\right] = \frac12$$

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