How to find the sum: $\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) $ $$S_n=\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) = \frac{1}{r-s}\left( \frac{1}{s+1}+\frac{1}{s+2}\cdots+\frac{1}{r}\right)$$
I fail to see how this:
$$S_n = \frac{1}{r-s}\left( \frac{1}{s+1}-\frac{1}{r-1}+\frac{1}{s+2}-\frac{1}{r+2}+\cdots\right)$$
converges to the right answer. Can someone please give a detailed explanation?
$r,s$ are integers and $r>s>0$

Consider the partial sum 
$$S_p=\sum_{n=0}^p  \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right)=\frac{1}{r-s}\left(H_{p+s+1}-H_{p+r+1}+H_r-H_s \right)$$ and use the asymptotics of harmonic numbers to get
$$S_p=\frac{H_r-H_s}{r-s}-\frac{1}{p}+\frac{r+s+3}{2
   p^2}+O\left(\frac{1}{p^3}\right)$$ Then, the limit.
