Find the value of $ \frac{1}{{2(2^2 - 1)}} + \frac{1}{{3(3^2 - 1)}} + \frac{1}{{4(4^2 - 1)}} + \cdots \;. $ To find the value of $$
\frac{1}{{2(2^2  - 1)}} + \frac{1}{{3(3^2  - 1)}} + \frac{1}{{4(4^2  - 1)}} +  \cdots \;.
$$
I presented it as $$
\sum\limits_{n = 2}^\infty  {\frac{1}{{n(n^2  - 1)}}} .
$$
Then using partial fractions method
I wrote the sum as
$$
\sum\limits_{n = 2}^\infty  {\left( { - \frac{1}{n} + \frac{1}{{2(n + 1)}} + \frac{1}{{2(n - 1)}}} \right)} .
$$
Then I wrote down the individual terms to see if most of the terms cancel out like they do sometimes.
$$
 - \frac{1}{2} + \frac{1}{6} + \frac{1}{2} - \frac{1}{3} + \frac{1}{8} + \frac{1}{4} - \frac{1}{4} + \frac{1}{{10}} + \frac{1}{6} - \frac{1}{5} + \frac{1}{{12}} + \frac{1}{8} - \frac{1}{6} + \frac{1}{{14}} + \frac{1}{{10}} +  \cdots \; .
$$
Unfortunately they don't seem to cancel out. Is there a way to find out the sum in this method or any other method?
 A: You can continue with the partial sums as
\begin{align*}
\sum\limits_{n = 2}^N {\left( { - \frac{1}{n} + \frac{1}{{2(n + 1)}} + \frac{1}{{2(n - 1)}}} \right)} & = \sum\limits_{n = 2}^N {\left( {\left( {\frac{1}{{2(n - 1)}} - \frac{1}{{2n}}} \right) + \left( {\frac{1}{{2(n + 1)}} - \frac{1}{{2n}}} \right)} \right)} \\  & =
\sum\limits_{n = 2}^N {\left( {\frac{1}{{2(n - 1)}} - \frac{1}{{2n}}} \right)}  + \sum\limits_{n = 2}^N {\left( {\frac{1}{{2(n + 1)}} - \frac{1}{{2n}}} \right)} \\
& = \frac{1}{4}- \frac{1}{{2N}}  + \frac{1}{{2(N + 1)}}  = \frac{1}{4} - \frac{1}{{2N(N + 1)}}.
\end{align*}
Now take the limit $N\to +\infty$ and conclude.
A: 
Note : $$\frac{1}{a.b.c}=\frac{1}{c-a}\left(\frac{1}{a.b}-\frac{1}{b.c}\right)$$
and $$\frac{1}{a.b}=\frac{1}{b-a}\left(\frac{1}{a}-\frac{1}{b}\right)$$

Use this in $$\sum_{n\geq2}\frac{1}{(n-1)(n)(n+1)}$$
Here's what you'll get : $$\sum_{n\geq2}\frac{1}{2}\left(\frac{1}{(n-1)(n)}-\frac{1}{(n)(n+1)}\right)\implies\frac{1}{2}\left(\color{red}{\sum_{n\geq2}\frac{1}{(n-1)(n)}}-\color{blue}{\sum_{n\geq2}\frac{1}{(n)(n+1)}}\right)\tag{*}$$
So , the first sum is $$\color{red}{\sum_{n\geq2}\frac{1}{(n-1)(n)}}=\sum_{n\geq2}\left(\frac{1}{(n-1)}-\frac{1}{n}\right)=1\tag1$$
and the secong sum will be $$\color{blue}{\sum_{n\geq2}\frac{1}{(n)(n+1)}}=\sum_{n\geq2}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1}{2}\tag2$$
Using the values of sums from $(1)$ and $(2)$ in $(*)$ $$\sum_{n\geq2}\frac{1}{n(n^2-1)}=\frac{1}{2}\left(\color{red}{1}-\color{blue}{\frac{1}{2}}\right)\implies\frac{1}{4}$$
