# 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...

## 3 Answers

We have

\begin{align} \sum_{k\ge1}\left(\frac{1}{k}-\frac{1}{k+m}\right)&=\sum_{k\ge1}\frac{1}{k}-\sum_{k\ge1}\frac{1}{k+m}\\ &=\sum_{k\ge1}\frac{1}{k}-\sum_{k+m\ge m+1}\frac{1}{k+m}\\ &=\sum_{k\ge1}\frac{1}{k}-\sum_{k\ge m+1}\frac{1}{k}\\ &=\sum_{k=1}^m\frac{1}{k}\\ &=H_m \end{align}

$$\require{cancel} \sum_{k\geq 1}\frac{1}{k}-\frac{1}{k+m} = \left(\frac{1}{1} - \cancel{\frac{1}{m+1}}\right) + \left(\frac{1}{2} - \cancel{\frac{1}{m+2}}\right) + \ldots + \left(\cancel{\frac{1}{m+1}} - \frac{1}{2m+1}\right) + \left(\cancel{\frac{1}{m+2}} - \frac{1}{2m+2} \right) + ...$$

So, all the second terms will cancel out and all the first terms after $k\geq m+1$ will also cancel out leaving,

$$\sum\limits_{k=1}^{m}\frac{1}{k}$$

Easier if you break down the first series:

$$=\sum_{k=1}^{\infty}\frac1k-\sum_{k=1}^{\infty}\frac{1}{k+m}=\sum_{k=1}^{m}\frac1k+\require{cancel}\cancel{\sum_{k=m+1}^{\infty}\frac1k}-\cancel{\sum_{k=1}^{\infty}\frac{1}{k+m}}=H_m$$