Does $\sum_{n=2}^\infty \frac{n^2-4}{(n-1)^2(n+3)^2} $ converge? Does $\sum_{n=2}^\infty \frac{n^2-4}{(n-1)^2(n+3)^2} $ converge?
I used the integral test and found that it does, but it was a bit cumbersome. Is there an easier way?
Thanks
 A: You can look for a bound. For example
$$0\le\frac{n^2-4}{(n-1)^2(n+3)^2}<\frac{n^2}{(n-1)^2n^2}=\frac{1}{(n-1)^2}$$
And it is simple to demonstrate that 
$$S=\sum_{n=2}^{\infty} \frac{1}{(n-1)^2}$$
converges, and the sum will be bounded between 
$$0\le\sum_{n=2}^\infty \frac{n^2-4}{(n-1)^2(n+3)^2}\le S$$
A: The easiest is to use equivalence, since it is a series with positive terms:
$$\frac{n^2-4}{(n-1)^2(n+3)^2}\sim_\infty\frac{n^2}{n^2\cdot n^2}=\frac1{n^2}, $$
which converge.
A: By partial fraction decomposition,
$$\begin{eqnarray*}\sum_{n\geq 3}\frac{n^2-2n-3}{(n-2)^2 (n+2)^2}&=&\frac{1}{32}\sum_{n\geq 3}\left(\color{blue}{\frac{7}{x-2}-\frac{7}{x+2}}+\color{red}{\frac{10}{(x+2)^2}-\frac{6}{(x-2)^2}}\right)\\&=&\frac{1}{32}\left[\color{blue}{\frac{175}{12}}\color{red}{-6\,\zeta(2)+10\left(\zeta(2)-\frac{1}{1^2}-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}\right)}\right]\\&=&\frac{1}{32}\left[\frac{25}{72}+4\,\zeta(2)\right]=\color{purple}{\frac{25}{2304}+\frac{\pi^2}{48}}.\end{eqnarray*}$$
The blue sum is a telescopic sum, the red sum is a well-known sum.
