$S = \sum_{m=1}^{\infty} \frac{1}{2m+s}\frac{1}{2m+r}$
$r>s>0$ and both $r,s$ integers.
write $x=2m+s$ and $y=r-s$. \begin{align} &S = \sum_{m=1}^{\infty} \frac{1}{2m+s}\frac{1}{2m+r} = \sum_{x=s+2}^{\infty} \frac{1}{x}\frac{1}{x+y} = \sum_{x=s+2}^{\infty} \left( \frac{1}{xy} - \frac{1}{y(x+y)} \right)=\\ &\frac{1}{y}\sum_{x=s+2}^{\infty} \left( \frac{1}{x} - \frac{1}{x+y} \right) = \frac{1}{r-s}\sum_{x=s+2}^{\infty} \left( \frac{1}{x} - \frac{1}{x+r-s} \right) \end{align}
From this post Evaluating the sum (telescoping series): $\sum_{k=s+1}^{\infty}\left( \frac{1}{k} - \frac{1}{k+r-s}\right)$.
\begin{align} S = \frac{1}{r-s}\sum_{x=s+2}^{\infty} \left( \frac{1}{x} - \frac{1}{x+r-s} \right) =\frac{1}{r-s} \left( \frac{1}{s+2} + \frac{1}{s+3} \cdots + \frac{1}{r}\right) \end{align}
Is the above correct? I asked this another day but I could understand nothing of the answers.
Another thing is: I assumed that both $r,s$ are integers, but, If I assumed to be both odd numbers, would something change?
I did some tests in Wolfram, and it seems that for both $r,s$ odd integers and $r>s>0$, the sum $S$ will be a rational number.
This is a follow up question to: Find the sum of $\sum_{k=0}^{\infty} 1/[(r+2k+2)(s+2k+2)]$
The answer in the link use the Digamma function. I wanted to know If the answer in the current post is correct.