Hint: Write $$\frac1{k(k+p)}=\frac{A}{k}+\frac{B}{k+p}.$$ Solve for $A$ and $B,$ and you'll find that the series does telescope.
Added: Now that I'm back at my computer and have a little time, and now that you've accepted an answer, I'll expand on my own. You should readily find that $A=\frac1p$ and $B=-\frac1p.$ Consequently, we can rewrite the series as $$\frac1p\sum_{k=1}^\infty\left(\frac1k-\frac1{k+p}\right).$$ To prove convergence and find the limit, it will suffice to consider the partial sums $$S_n:=\frac1p\sum_{k=1}^n\left(\frac1k-\frac1{k+p}\right).$$ We'd like to get this into a more convenient form, as a difference of sums, rather than a sum of differences. That is, we'll rewrite it as $$S_n=\frac1p\sum_{k=1}^n\frac1k-\frac1p\sum_{k=1}^n\frac1{k+p},\tag{1}$$ which identity is readily proved by arithmetic properties.
Now, take any integer $m\ge1$ and note that $$\sum_{k=1}^{p+m}\frac1k=\sum_{k=1}^p\frac1k+\sum_{k=p+1}^{p+m}\frac1k.\tag{2}$$ On the other hand, $$\sum_{k=1}^{p+m}\frac1{k+p}=\sum_{k=1}^{p+m}\frac1{p+k}=\sum_{k=1}^m\frac1{p+k}+\sum_{k=m+1}^{p+m}\frac1{p+k}=\sum_{k=p+1}^{p+m}\frac1k+\sum_{k=m+1}^{p+m}\frac1{p+k}.\tag{3}$$
Consequently, by $(1)$ through $(3),$ we have for any integer $m\ge1$ that $$S_{p+m}=\frac1p\sum_{k=1}^p\frac1k-\frac1p\sum_{k=m+1}^{p+m}\frac1{p+k}=\frac1p\sum_{k=1}^p\frac1k-\frac1p\sum_{k=1}^p\frac1{p+m+k}.\tag{$\star$}$$ Now, for any such integer $m,$ we have $$0<\sum_{k=1}^p\frac1{p+m+k}<\sum_{k=1}^p\frac1{p+m+1}=\frac{p}{p+m+1},$$ so $$-\frac1{p+m+1}<-\frac1p\sum_{k=1}^p\frac1{p+m+1}<0,$$ and so by $(\star),$ we have $$-\frac1{p+m+1}<S_{p+m}-\frac1p\sum_{k=1}^p\frac1k<0$$ for all integers $m\ge1.$ A quick application of the Squeeze Theorem shows that the sequence of partial sums converges to $\frac1p\sum\limits_{k=1}^p\frac1k,$ proving series convergence and giving us its sum.
Best Answer
$$\frac{1}{k(k+2)} = \dfrac12\left(\frac{1}{k}-\dfrac{1}{k+1}\right)+\dfrac12\left(\dfrac{1}{k+1}-\frac{1}{k+2}\right)$$