Obtaining the sum of series using the partial fraction method.

Question

Let

$\lim_{n\rightarrow \infty }\sum_{k=2}^{n}\frac{1}{k^3-k}$

I converted the above series into following form by partial fraction method:

$\sum_{k=2}^{n}\frac{1}{k^3-k} =\sum_{k=2}^{n}\frac{-1}{k}+\frac{1}{2} \sum_{k=2}^{n}\frac{1}{(k+1)(k-1)}$

The sum of the second series can be done using the following method: $\frac{1}{4}\sum_{k=2}^{n}\frac{(k+1)-(k-1)}{(k+1)(k-1)}=\frac{1}{4}\sum_{k=2}^{n}[\frac{1}{k-1}-\frac{1}{k+1}]$

Which, then can be easily solved. But the first series clearly a divergent series. How can I then be able to solve the problem? The answer is given as one fourth.

Answer 1

$$\frac{1}{k^3-k}=\frac{1}{(k-1)k(k+1)}=\frac{1}{2}\left[\frac{1}{(k-1)k}-\frac{1}{k(k+1)}\right] $$ which is a telescopic term, hence $$ \sum_{k=2}^{n}\frac{1}{k^3-k}=\frac{1}{2}\left[\frac{1}{(2-1)2}-\frac{1}{n(n+1)}\right] $$ and the limit as $n\to +\infty$ is $\frac{1}{4}$.

Answer 2

$$\lim_{n\rightarrow \infty }\sum_{k=2}^{n}\frac{1}{k^3-k}$$
$$ =\lim_{n\rightarrow \infty }\sum_{k=2}^{n}\frac{1}{k (k+1)(k-1)}$$
$$ =\lim_{n\rightarrow \infty }\sum_{k=2}^{n}\frac{(k+1)-(k-1)}{k (k+1)(k-1)}$$
$$ =\lim_{n\rightarrow \infty }\frac{1}{2}\left[\frac{1}{(k-1)k}-\frac{1}{k(k+1)}\right]$$
$$= \lim_{n\rightarrow \infty }\frac{1}{2}\left[\frac{1}{(2-1)2}-\frac{1}{n(n+1)}\right]= \frac{1}{4}$$