$$ \sum_{n=4}^\infty \frac{2}{(n-1)(n-2)(n)} $$ is a telescoping series but how can I make it more visual? How can I write that in telescoping form? $$\sum_{n=1}^\infty(a_n - a_{n+1})$$
I already have then in partial fractions,
$$ \sum_{n=4}^\infty \left(\frac{1}{n-2} -\frac{2}{n-1} + \frac{1}{n} \right) $$
We can re-write, $$\left(\frac{2}{n-1}\right)=\left(\frac{1}{n-1} + \frac{1}{n-1}\right)$$
and split in two telescoping-series:
$$ \sum_{n=4}^\infty \left(\frac{1}{n-2} +\frac{1}{n-1} \right) - \sum_{n=4}^\infty \left(\frac{1}{n-1} - \frac{1}{n}\right) $$
so the formula for the partial sum will be:
$$S_n = \left(\frac{1}{2} - \frac{1}{3} - \frac{1}{n-1} - \frac{1}{n}\right)$$
and the lim of $S_n$ will give us the convergence:
$$\lim_{n\to\infty} S_n = \frac{1}{6}$$
I would like to thank everyone who responded my question
In order to determine the value the series converges to, manipulate the terms as follows: $$\sum_{n=4}^{\infty}\frac{1}{n-1}=\sum_{n=3}^{\infty}\frac{1}{n}$$ $$\sum_{n=4}^{\infty}\frac{1}{n-2}=\sum_{n=2}^{\infty}\frac{1}{n}=\frac{1}{2}+\sum_{n=3}^{\infty}\frac{1}{n}$$
It should converge to $1/6$.
Allow me to revise: \begin{align} \sum_{n=4}^N\left(\frac{1}{n-2}-\frac{2}{n-1}+\frac{1}{n}\right) &=\sum_{n=4}^N\frac{1}{n-2}-2\sum_{n=4}^N\frac{1}{n-1}+\sum_{n=4}^N\frac{1}{n}\\ &=\sum_{n=2}^{N-2}\frac{1}{n}-2\sum_{n=3}^{N-1}\frac{1}{n}+\sum_{n=4}^{N}\frac{1}{n}\\ &=\left(\frac{1}{2}+\frac{1}{3}+\sum_{n=4}^{N-2}\frac{1}{n}\right)-\left(\frac{2}{3}+\frac{2}{N-1}+2\sum_{n=4}^{N-2}\frac{1}{n}\right)+\left(\frac{1}{N-1}+\sum_{n=4}^{N-2}\frac{1}{n}\right)\\ &=\frac{1}{2}+\frac{1}{3}-\frac{2}{3}-\frac{1}{N-1}\\ &=\frac{1}{6}-\frac{1}{N-1} \end{align} As $N\to\infty$, it is now clear that the sum converges to $1/6$.