# infinity series of Riemann zeta function at odd integers

Properties of Riemann zeta function at odd and even integers diverge dramatically, which can be proved by many evidences.

I once found an infinity series in wikipedia, it reads $$\sum_{n=1}^{\infty}\frac{\zeta(2n)-1}{a^{2n}} = \frac12+\frac{1}{1-a^2}-\frac{\pi\cot(\pi/a)}{2a},~\vert a\vert>1$$ or equivalently, $$\sum_{n=1}^{\infty}\frac{\zeta(2n)}{a^{2n}} = \frac12-\frac{\pi\cot(\pi/a)}{2a},~\vert a\vert>1.$$

Here, I just wonder that, if there is a closed form for the series $$I(a)=\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{a^{2n+1}} = \mathbf{?},~\vert a\vert>1.$$ Thanks a lot. Any suggestion or material link will be welcomed.

EDIT: As @Lucian's hint in the comment, I arrive at $$I(a)=\sum_{n=1}^{\infty}\frac{1}{a^{2n+1}}\Big(\sum_{k=1}^{\infty}\frac{1}{k^{2n+1}}\Big)=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{(ka)^{2n+1}}=\sum_{k=1}^{\infty}\frac{1}{(ka)[(ka)^2-1]}$$ However, the trick $$\frac{1}{x(x^2-1)}=\frac{x}{2}\Big(\frac{1}{x-1}-\frac{1}{x+1}\Big)-\frac{1}{x}$$ doesn't seem to work for the above series. What should I do now?

• Hint: Expand the $\zeta$ function into its infinite series, and then switch the order of the two summations. Then read this. – Lucian May 28 '15 at 12:41
• See my edit of my question. What should I do next? No closed form appears. – Roger209 May 28 '15 at 13:04
• I've never tried this, but I assume it might be difficult because of our lack of knowledge of the zeta function at odd integers. That is, a closed form value – ClassicStyle May 28 '15 at 13:04
• Further Hint: Differentiate the natural logarithm of Euler's infinite product expression for the sine function. Then take a better look at what you've obtained so far. – Lucian May 28 '15 at 13:08
• @Lucian: Yes, I did. Differentinating "\ln(\sin x)" from the Euler product, I find the formula helps me calculate the series related to the EVEN integers, not the odd ones. Joining TylerHG 's comment, it seems that no closed form exists for the odd integers. Is it? – Roger209 May 28 '15 at 13:38

## 1 Answer

There exists a closed form in terms of special functions. We have $$\sum_{k\geq1}\frac{1}{ka\left(\left(ka\right)^{2}-1\right)}=\frac{1}{a^{3}}\sum_{k\geq1}\frac{1}{k\left(k^{2}-\frac{1}{a^{2}}\right)}=\frac{1}{a^{3}}\sum_{k\geq1}\frac{1}{k\left(k-\frac{1}{a}\right)\left(k+\frac{1}{a}\right)}$$ and using the identity $$\psi\left(1+z\right)=-\gamma+\sum_{k\geq1}\frac{z}{k\left(k+z\right)}$$ where $\psi\left(z\right)$ is the digamma function and $z\notin\mathbb{Z}^{-}\setminus\left\{ 0\right\}$, we have $$\sum_{k\geq1}\frac{1}{ka\left(\left(ka\right)^{2}-1\right)}=-\frac{\psi\left(1+\frac{1}{a}\right)+\psi\left(1-\frac{1}{a}\right)+2\gamma}{2a}.$$

• Thanks a lot. By virtue of $\psi(1+x)=\psi(x)+1/x$ and $\psi(1-x)=\psi(x)+\pi\cot(\pi x)$, the above formula can be more compact. – Roger209 May 29 '15 at 5:25