Show $\frac1{k-1}+\frac 6k-\frac{7}{k+1}=\frac{8k-6}{k(k^2-1)}$, deduce $\sum\limits_{k = 2}^n\frac{4k-3}{k(k^2-1)}$
If $k > 1$, then $\frac1{(k - 1)^2}-\frac1{(k + 1)^2}=\frac{4k}{(k^2 - 1)^2}$, hence simplify $\sum\limits_{k = 2}^n\frac k{(k^2 - 1)^2}$