How to write this sum $\sum_{m=0}^\infty (2m+1)^{-2k-1} \sum_{r=1}^\infty (-1)^{r-1} e^{-2(2m+1)r a} $ as a sum over single index?

Question

So I want to write the sum $$\sum_{m=0}^\infty (2m+1)^{-2k-1} \sum_{r=1}^\infty (-1)^{r-1} e^{-2(2m+1)r a} $$ where $a>0$ and $k\in \mathbb{N}$, as a sum over single index which probably uses odd divisors of $(2m+1)\cdot r$.

However, I am quite confused about this method in general. I am trying to learn this concept. If anyone can help me with the logic, it will be highly appreciated.

Answer 1

Let $$ S := \sum_{m=0}^\infty (2m+1)^{-2k-1} \sum_{r=1}^\infty (-1)^{r-1} e^{-2(2m+1)r a}. $$ Then, as you suggested using odd divisors, get $$ S = \sum_{n=1}^\infty (-1)^{n-1} e^{-2\,n\,a}\left( \sum_{d|n,\,d\text{ odd}} d^{-2k-1} \right) $$ where $\,d=2m+1\,$ and $\,n=(2m+1)r.$

Answer 2

I'm not sure if you're supposed to do this some specific way. I would just utilize the fact that the inner sum is a geometric series. You can easily get $$\sum\limits_{r=1}^{\infty}(-1)^{r-1}e^{-2(2m+1)ra} = -\sum\limits_{r=1}^{\infty}\left(-e^{-2(2m+1)a}\right)^{r} = \begin{cases}\frac{e^{-2(2m+1)a}}{1+e^{-2(2m+1)a}} & a > 0\\ \infty & \text{otherwise}\end{cases}.$$ This can be rewritten as $$\sum\limits_{r=1}^{\infty}(-1)^{r-1}e^{-2(2m+1)ra} = \begin{cases}\frac{1}{1+e^{2(2m+1)a}} & a > 0\\ \infty & \text{otherwise}\end{cases}.$$ I don't know if this helps. I hope it does!