Show the following: $\displaystyle\sum_{n=1}^\infty\dfrac1{n(n+k)}=\dfrac{H_k}k$.

Question

For each $n\in\Bbb N$ with $n\geq1$ is $\displaystyle H_n:=\sum_{k=1} ^n\dfrac 1k$ the $n$*-th partial sum of the harmonic series.* $k\in\Bbb N$ with $k\geq1$. Show that $\displaystyle\sum_{n=1}^\infty\dfrac1{n(n+k)}=\dfrac{H_k}k$

How would I go about soving this? I've been doing math all day long and my head is about to explode. But I need this for tomorrow. Help please.

Answer 1

Note that $$\frac{1}{n(n+k)} = \frac{1}{k}\left(\frac{1}{n} - \frac{1}{n+k}\right).$$