Prove that the series $\sum_{1}^{\infty}\frac{k}{(k+1)(k+2)(k+3)}$ converges and find its limit I try to split the summand into differences, but that seems to be a futile way in our case right here, because the numerator is $k$, instead of a given number. 
A closely-related series, say $\sum_{1}^{\infty}\frac{2}{(k+1)(k+2)(k+3)}$, can however be directly tackled by writing its partial sum as
$$\sum_{1}^{n}\frac{1}{(k+1)(k+2)} - \frac{1}{(k+2)(k+3)}.$$

Since:
$$-\frac{1}{k+1}+\frac{4}{k+2}-\frac{3}{k+3}=\frac{2k}{(k+1)(k+2)(k+3)}$$
we have:
$$S_N=\sum_{n=1}^{N}\frac{k}{(k+1)(k+2)(k+3)}=\sum_{k=1}^{N}\left(\frac{-1/2}{k+1}+\frac{2}{k+2}+\frac{-3/2}{k+3}\right) $$
and since $(-1/2)+(2)+(-3/2)=0$ we have:
$$ \lim_{N\to +\infty}S_N = \frac{-1/2}{1+1}+\frac{2}{1+2}+\frac{-1/2}{2+1}=\color{red}{\frac{1}{4}}.$$
The convergence of the original series is trivial since:
$$0\leq \frac{k}{(k+1)(k+2)(k+3)}\leq\frac{1}{(k+2)(k+3)}$$
and:
$$\sum_{k\geq 1}\frac{1}{(k+2)(k+3)}=\frac{1}{3}.$$
