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 $r>s>0$
 A: Since $r > s$,
$$\begin{array}{rl} & \frac{1}{n+s+1} & - \frac{1}{n+r+1}\\
= & \frac{1}{n+s+1} &+ \color{red}{\frac{1}{n+s+2}} &+ \cdots &+ \color{blue}{\frac{1}{n+r}}\\
  & & - \color{red}{\frac{1}{n+s+2}} &- \cdots &- \color{blue}{\frac{1}{n+r}} &- \frac{1}{n+r+1}\end{array}$$
we have
$$\begin{align}\frac{1}{n+s+1} - \frac{1}{n+r+1} 
&= \sum_{k=1}^{r-s} \frac{1}{n+s+k} - \sum_{k=1}^{r-s} \frac{1}{(n+1)+s+k}\\
&= U(n) - U(n+1)\end{align}
$$
where $U(n) = \sum\limits_{k=1}^{r-s}\frac{1}{n+s+k}$. The sum at hand is a telescoping sum. The partial sums has the form
$$\begin{align}
\sum_{n=0}^p \frac{1}{r-s}\left(\frac{1}{n+s+1} - \frac{1}{n+r+1}\right)
= & \frac{1}{r-s}\sum_{n=0}^p \left(U(n) - U(n+1)\right)\\
= & \frac{1}{r-s}\left(U(0) - U(p+1)\right)\end{align}$$
Since $U(p+1)$ is a finite sum of $r-s$ terms and each terms converge to $0$ as $p\to \infty$, we find
$$\lim_{p\to\infty} U(p+1) = 0$$
As a result,
$$\begin{align}
& \sum_{n=0}^\infty \frac{1}{r-s}\left(\frac{1}{n+s+1} - \frac{1}{n+r+1}\right) \\
= & \lim_{p\to\infty} \sum_{n=0}^p \frac{1}{r-s}\left(\frac{1}{n+s+1} - \frac{1}{n+r+1}\right)\\
= & \lim_{p\to\infty}\frac{1}{r-s}\left(U(0) - U(p+1)\right)\\
= & \frac{1}{r-s}U(0)\\
= & \frac{1}{r-s}\sum_{k=1}^{r-s}\frac{1}{s+k}\\
= & \frac{1}{r-s}\left(\frac{1}{s+1} + \frac{1}{s+2} + \cdots + \frac{1}{r}\right)
\end{align}$$
A: Let $r=s+k$. Then:
$$\begin{align}S_n&=\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) = \\
&=\frac1{r-s}(\ \boxed{\color{black}{\frac1{s+1}}}-\color{red}{\frac1{s+k+1}}+\color{black}{\boxed{\frac1{s+2}}}-\color{blue}{\frac1{s+k+2}}\color{black}{+\cdots +\boxed{\frac1{s+k}}}-\color{green}{\frac1{s+2k}}+\\
&\ \  \ \  +\color{red}{\frac1{s+k+1}}-\color{#FF00FF}{\frac1{s+2k+1}}+\color{blue}{\frac1{s+k+2}}-\color{#6495ED}{\frac1{s+2k+2}}+\cdots +\color{green}{\frac1{s+2k}}-\color{red}{\frac1{s+3k}}+\\
&\ \  \ \  +\color{#FF00FF}{\frac1{s+2k+1}}\color{black}{-\frac1{s+3k+1}}+\color{#6495ED}{\frac1{s+2k+2}}\color{black}{-\frac1{s+3k+2}+\cdots +}\color{red}{\frac1{s+3k}}\color{black}{-\frac1{s+4k}+\cdots})=\\
&=\frac{1}{r-s}\left( \frac{1}{s+1}+\frac{1}{s+2}\cdots+\frac{1}{s+k}\right)=\\
&=\frac{1}{r-s}\left( \frac{1}{s+1}+\frac{1}{s+2}\cdots+\frac{1}{r}\right).\end{align}$$
A: 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.
