<s>[INST] How to compute the series $\sum_{k\geq 1}\frac{1}{k(k+m)}$ I tried to do it like this:
$$\sum_{k\geq 1} \frac{1}{k(k+m)}=$$ $$\ = \sum_{k\geq 1}  \frac{1}{km}-\frac{1}{m(k+m)}=$$ $$=\sum_{k\geq 1}  \frac{1}{m}(\frac{1}{k}-\frac{1}{k+m})=$$ $$=\frac{1}{m}\sum_{k\geq 1}\frac{1}{k}-\frac{1}{k+m} $$
I do know that this sum should equal to$\ \frac{H_m}{m}$, but i don't know how to prove it from here...
 [/INST] $$\require{cancel} \sum_{k\geq 1}\frac{1}{k}-\frac{1}{k+m} = \left(\frac{1}{1} - \cancel{\frac{1}{m+1}}\right) + \left(\frac{1}{2} - \cancel{\frac{1}{m+2}}\right) + \ldots + \left(\cancel{\frac{1}{m+1}} - \frac{1}{2m+1}\right) + \left(\cancel{\frac{1}{m+2}} - \frac{1}{2m+2} \right) + ...$$
So, all the second terms will cancel out and all the first terms after $k\geq m+1$ will also cancel out leaving,
$$\sum\limits_{k=1}^{m}\frac{1}{k}$$
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