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?
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 $$\sum_{k\geq1}\frac{1}{ka\left(\left(ka\right)^{2}-1\right)}=-\frac{\psi\left(1+\frac{1}{a}\right)+\psi\left(1-\frac{1}{a}\right)+2\gamma}{2a}. $$