<s>[INST] Convergence of $ \sum\limits_{n=1}^{\infty}\frac{1}{n(n+m)} $ Let $ m \in \mathbb{N} $. Find to what does the series $ \sum\limits_{n=1}^{\infty}\frac{1}{n(n+m)}  $ converge to
My attempt: ( I did as discussed in Infinite Series $\sum 1/(n(n+1))$ )
Let $ N>m $.
$ 
\begin{align}
S_N & = \sum\limits_{n=1}^{N}\frac{1}{n(n+m)}  = \frac{1}{m}\sum_{n=1}^N \left(\dfrac1n - \dfrac1{n+m}\right) = \frac{1}{m}( \sum_{n=1}^N \dfrac1 n - \sum_{n=1}^N \dfrac1{n+m}) = \frac{1}{m}(\sum_{n=1}^N \dfrac1n - \sum_{n=m+1}^{N+m} \dfrac1{n} )\\
& = \frac{1}{m}(\sum_{n=1}^m \dfrac1n + \sum_{n=m+1}^N \dfrac1n - \sum_{n=m+1}^N\dfrac1n -  \sum_{n=N+1}^{N+m}\dfrac1n ) = \frac{1}{m}(\sum_{n=1}^m \dfrac1n  -  \sum_{n=N+1}^{N+m}\dfrac1n ) 
~~\text{ [ from here I got stuck] } \end{align} 
 $
 [/INST] You are almost done.  We have $$\begin{align}
S_N &= \frac{1}{m} \sum_{n=1}^m \frac{1}{n} - \frac{1}{m} \sum_{n = N+1}^{N+m} \frac{1}{n} \\
&= \frac{1}{m} \sum_{n=1}^m \frac{1}{n} - \frac{1}{m} \sum_{n = 1}^{m} \frac{1}{n+N} \\
&= \frac{1}{m} H_m - \frac{1}{m} \sum_{n = 1}^{m} \frac{1}{n+N},
\end{align}$$
where I have used $H_m = \sum_{n=1}^m \frac{1}{n}$ to represent the $m^{\rm th}$ harmonic number.  hence for a fixed $m$, as $N \to \infty$,
$$\lim_{N \to \infty} S_N = \frac{H_m}{m} - \frac{1}{m} \lim_{N \to \infty} \sum_{n=1}^m \frac{1}{n + N}.$$  But since the remaining sum contains only finitely many terms, we can interchange the order of summation and limit to obtain
$$\lim_{N \to \infty} S_N = \frac{H_m}{m} - \frac{1}{m}  \sum_{n=1}^m \lim_{N \to \infty} \frac{1}{n + N} = \frac{H_m}{m} - \frac{1}{m} \sum_{n=1}^m 0 = \frac{H_m}{m}.$$
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