Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Isn't there an exact expression for the zeta function of even integers? – Ron Gordon Feb 15 '13 at 12:20
@rlgordonma, I m not familiar with zeta :(. – Ryan Feb 15 '13 at 12:21
May first idea is partial fraction decomposition – Dominic Michaelis Feb 15 '13 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. – Shobhit Feb 15 '13 at 13:24
@ShobhitBhatnagar That's quite a voluminous comment. Maybe this would be better in the answer zone. – 1015 Feb 15 '13 at 13:30
up vote 8 down vote accepted

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 $$ S=\tfrac12. $$

share|cite|improve this answer
Cool, nice to have something rigorous! – achille hui Feb 15 '13 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: $$\sum_{k=1}^{\infty} \frac{\zeta(2k)}{(k+1)(2k+1)} = \int_0^{\infty} \left[\frac{1+e^{-x}}{x^2} - 2 \frac{1-e^{-x}}{x^3}\right] dx = \int_0^{\infty} \frac{d}{dx}\left[-\frac{1}{x} + \frac{1-e^{-x}}{x^2}\right]dx\\ = -\lim_{x\to 0} \left[-\frac{1}{x} + \frac{1-e^{-x}}{x^2}\right] = \frac12 $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.