If $k > 1$, then $\frac1{(k - 1)^2}-\frac1{(k + 1)^2}=\frac{4k}{(k^2 - 1)^2}$, hence simplify $\sum\limits_{k = 2}^n\frac k{(k^2 - 1)^2}$ I have the following problem:

Prove that if $k > 1$, then
$$\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{4k}{(k^2 - 1)^2}$$
Hence simplify
$$\sum\limits_{k = 2}^n \dfrac{k}{(k^2 - 1)^2}$$

My Work:
$$\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{(k + 1)^2 - (k - 1)^2}{(k - 1)^2 (k + 1)^2} = \dfrac{(k + 1)^2 - (k - 1)^2}{(k^2 - 2k + 1)(k^2 + 2k + 1)}$$
I could have continued and multiplied the denominator out, but, at this point, I'm thinking that I'm actually expected to have proceeded differently, rather than multiplying the denominator out fully? And I don't have solutions for this problem, so I can't check anything.
I would appreciate it if someone could please take the time to explain this problem.

Here are some guidelines:


*

*Expand $(k+1)^2-(k-1)^2$, or use the rule $a^2-b^2=(a-b)(a+b)$.

*Notice that $(k-1)(k+1)=k^2-1$, thus $(k-1)^2(k+1)^2=?$

*Once you show that $$\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{4k}{(k^2 - 1)^2},$$ you can use it so that the sum becomes\begin{align*}
\sum_{k=2}^n\dfrac{k}{(k^2-1)^2}&=\dfrac{1}{4}\sum_{k=2}^n\left(\dfrac{1}{(k-1)^2}-\dfrac{1}{(k+1)^2}\right)\\
&=\dfrac{1}{4}\left(\sum_{k=2}^n\dfrac{1}{(k-1)^2}-\sum_{k=2}^n\frac{1}{(k+1)^2}\right);
\end{align*}
notice that by changing the variable $k-1$ into $k$, you get $$\sum_{k=2}^n\dfrac{1}{(k-1)^2}=\sum_{k=1}^{n-1}\dfrac{1}{k^2}.$$ Do the same for $\sum_{k=2}^n\frac{1}{(k+1)^2}$ and see what you'll get.
