# 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. May 28, 2015 at 12:41
• See my edit of my question. What should I do next? No closed form appears. May 28, 2015 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 May 28, 2015 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. May 28, 2015 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? May 28, 2015 at 13:38

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. May 29, 2015 at 5:25