I found the following identity:
For $r>s>0$ and integers. \begin{align} \sum_{m=1}^{\infty} \frac{1}{m+r} \frac{1}{m+s} = \frac{H_r-H_s}{r-s} = \frac{1}{r-s} \left( \frac{1}{s+1} + \frac{1}{s+2} + \cdots + \frac{1}{r}\right) \end{align} For $r>s>0$ and both $r,s$ odd integers. \begin{align} \sum_{m=1}^{\infty} \frac{1}{2m+r} \frac{1}{2m+s} = \frac{H_{r/2}-H_{s/2}}{2(r-s)} = \:\: ? \end{align}
What is the difference between $H_{r/2}-H_{s/2} \:\:?$
EDIT: I think the correct answer is: \begin{align} \sum_{m=1}^{\infty} \frac{1}{2m+r} \frac{1}{2m+s} = \frac{1}{r-s} \left( \frac{1}{s+2} + \frac{1}{s+4} + \cdots + \frac{1}{r} \right) \end{align}