Telescoping with a shift p

Convergence. $\frac1{k(k+p)}\le\frac1{k^2}$, which is summable, so the series converges by comparison.

Partial sums.
$$S_n=\frac1p\sum_{k=1}^n\left(\frac1k-\frac1{k+p}\right)=\frac1p\left(\sum_{k=1}^p\frac1k-\sum_{k=n+1}^{n+p}\frac1k\right).$$
The second group has $p$ terms, each tending to $0$.

Sum.
$$\sum_{k=1}^\infty\frac1{k(k+p)}=\frac1p\sum_{k=1}^p\frac1k=\frac{H_p}p.$$