Difficult Telescoping Series Finding the explicit sum of a telescoping series with two factors in the denominator is not a problem: we split the fractions in the difference of two pieces.
But what about 2+ factors
\begin{equation}
\lim_{n\to\infty} \sum \frac{1}{k(k+2)(k+4)}
\end{equation}
 A: hint: we can do telescoping inside teloscoping as follows:
$\dfrac{1}{n(n+2)(n+4)}= \dfrac{1}{4}\left(\dfrac{1}{n(n+2)} - \dfrac{1}{(n+2)(n+4)}\right)$, and split again: $\dfrac{1}{n(n+2)} = \dfrac{1}{2}\left(\left(\dfrac{1}{n} - \dfrac{1}{n+1}\right) + \left(\dfrac{1}{n+1} - \dfrac{1}{n+2}\right)\right)$. Do the same for the other term and there are a total of $4$ "mini" telescoping sums to be evaluated.
A: $$\frac{1}{k (k+2) (k+4)}=-\frac{1}{4 (k+2)}+\frac{1}{8 (k+4)}+\frac{1}{8 k}=\frac{1}{8} \left(\frac{1}{k}-\frac{2}{k+2}+\frac{1}{k+4}\right)=$$
$$=\frac{1}{8} \left[\left(\frac{1}{k}-\frac{1}{k+2}\right)-\left(\frac{1}{k+2}-\frac{1}{k+4}\right)\right]$$
So you have two telescoping series to work on. $S_n=$
$$=\frac18\left[\left(1-\frac13+\frac12-\frac13+\ldots+\frac{1}{k}-\frac{1}{k+2}\right)-\left(\frac13-\frac15+\frac14-\frac15+\ldots+\frac{1}{k+2}-\frac{1}{k+4}\right)\right]$$
$$=\frac18\left[\left(1+\frac12-\frac{1}{k+2}\right)-\left(\frac13+\frac14-\frac{1}{k+4}\right)\right]\to \frac{11}{96} \text{ as }n\to \infty$$
