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 have done series with $\zeta(2k)$ and $\zeta(k)$, but I have no idea with this one:

$$\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1}=-\gamma+\log(2)$$

$\gamma$ is the Euler–Mascheroni Constant.

This value was given by Mathematica. Any hint?

share|cite|improve this question
Expand $\zeta(2 k + 1)$ and interchange summation order? – vonbrand Apr 13 '13 at 14:06
@vonbrand I think you get a logarithmic sum by doing that. How do you evaluate that? – Ishan Banerjee Apr 13 '13 at 14:47
The same constant arises in $$lim_{n \to \infty} \left(log(2n+1)-H_n\right) = log(2)-\gamma$$ – Jaume Oliver Lafont Jan 10 at 23:20
up vote 18 down vote accepted

I solved it myself.

First we note that

$$\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1} = \sum_{n=2}^\infty \sum_{k=1}^\infty \frac{1}{(k+1)n^{2k+1}}=\sum_{n=2}^\infty \left( -\frac{1}{n}- n\log \left( 1-\frac{1}{n^2}\right)\right)$$

Then $$\begin{aligned} \sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1} &=\sum_{n=2}^\infty \left( -\frac{1}{n}- n\log \left( 1-\frac{1}{n^2}\right)\right) \\ &= \lim_{N\to \infty}\sum_{n=2}^N \left( -\frac{1}{n}- n\log \left( 1-\frac{1}{n^2}\right)\right)\\ &= \lim_{N\to \infty} \left[ -H_N+1-\sum_{n=2}^N n \log(n^2-1)+2\sum_{n=2}^Nn\log(n)\right]\\ &= \lim_{N\to \infty} \left[ -H_N+1-\sum_{n=2}^N \left(n\log(n+1) +n\log(n-1)-2n\log(n)\right)\right] \\ &= \lim_{N\to \infty} \Bigg[ -H_N+1+\log(2)-\sum_{n=3}^{N+1}(n-1)\log(n)-\sum_{n=3}^{N-1}(n+1)\log(n) \\ &\quad+\sum_{n=3}^N2n\log(n)\Bigg] \\ &= \lim_{N\to \infty}\left[-H_N-N\log(N+1)-(N-1)\log(N)+2N\log(N)+1+\log(2) \right]\\ &= \lim_{N\to \infty}\left(- \left(H_N-\log N \right)+\log(2)+1-N\log \left( 1+\frac{1}{N}\right)\right)\\&= \lim_{N\to \infty}\left( - \left(H_N-\log N \right)+\log(2)+\mathcal{O}(N^{-1})\right) \end{aligned}$$

Since $\displaystyle \gamma=\lim_{N\to \infty}(H_N-\log(N))$, we get

$$\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1} =-\gamma+\log(2)$$

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.