# Series involving the Riemann zeta function

Consider the series:

$$\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{n(2n+1)}$$

We can easily prove that it's a convergent series. My question, is there a way to express this series in terms of zeta constants ?

• What are zeta constants? – Igor Rivin Jan 16 '14 at 0:14
• A similar question has been asked recently. You have to replace the zeta function with its definition as an infinite series, then switch the order of summation. Also, use the fact that $\dfrac1{n(2n+1)}=\dfrac2{2n(2n+1)}=2\left(\dfrac1{2n}-\dfrac1{2n+1}\right)$ – Lucian Jan 16 '14 at 0:35
• @Lucian could you please refer me to that question ? – Mohammad Al Jamal Jan 16 '14 at 17:46
• I think it must have been one of these. – Lucian Jan 17 '14 at 6:11

## 1 Answer

It is not difficult to find the generating function for the values of the $\zeta$ function over the positive odd integers:

$$f(x)=\sum_{n\geq 1}\zeta(2n+1) x^{2n} = -\gamma-\frac{1}{2}\left[\psi(1-x)+\psi(1+x)\right] \tag{1}$$ and: $$\sum_{n\geq 1}\frac{\zeta(2n+1)}{n(2n+1)}=2\sum_{n\geq 1}\zeta(2n+1)\left(\frac{1}{2n}-\frac{1}{2n+1}\right)=2\int_{0}^{1}f(x)\left(\frac{1}{x}-1\right)\,dx\tag{2}$$

but the integrals $\int\frac{\psi(1\pm x)}{x}\,dx$ or $\int \log(x)\,\psi'(1\pm x)\,dx$ do not have nice closed forms (by my knowledge) but in terms of the Hurwitz zeta function and its derivatives.