How to find the value this sum converges to?$\sum_{n=2}^{\infty}\frac{n+1}{n(2n-1)(2n+1)} $ How to find the value this sum converges to?$$\sum_{n=2}^{\infty}\frac{n+1}{n(2n-1)(2n+1)} $$
I've tried separating it like
$$\sum_{n=2}^{\infty}\frac{1/2}{(2n+1)}+\frac{3/2}{(2n-1)}-\frac{1}{n}$$
and writing some terms and I get
$$1/6·(1+1/2+...+1/6n+3)+1/2·(1/3+1+1/2+...+1/6n-3)-(1+1/2+...1/n)$$
but I don't know how to end summing it all! Any hint!
FYI I haven't learnt integration and differentation.

Proceeding on the partial fraction decomposition you already got
$$
\eqalign{
  & S = \sum\limits_{2\, \le \,n} {{{n + 1} \over {n\left( {2n - 1} \right)\left( {2n + 1} \right)}}}  =   \cr 
  &  = \sum\limits_{2\, \le \,n} {{{1/2} \over {\left( {2n + 1} \right)}} + {{3/2} \over {\left( {2n - 1} \right)}} + {1 \over n}}  =   \cr 
  &  = \sum\limits_{0\, \le \,n} {{{1/2} \over {\left( {2n + 5} \right)}} + {{3/2} \over {\left( {2n + 3} \right)}} - {1 \over {n + 2}}}  =   \cr 
  &  = \sum\limits_{0\, \le \,n} {{{1/4} \over {\left( {n + 5/2} \right)}} + {{3/4} \over {\left( {n + 3/2} \right)}} - {1 \over {n + 2}}}  \cr} 
$$
Since
$$
\sum\nolimits_{n = 0}^N {{1 \over {n + a}}}  = \sum\nolimits_{n = a}^{N + a} {{1 \over n}}  = \psi (N + a) - \psi (a)
$$
then
$$
\eqalign{
  & S = \mathop {\lim }\limits_{N\, \to \,\infty } \left( \matrix{
  {1 \over 4}\psi (N + 5/2) + {3 \over 4}\psi (N + 3/2) - \psi (N + 2) +  \hfill \cr 
   + \psi (2) - {1 \over 4}\psi (5/2) - {3 \over 4}\psi (3/2) \hfill \cr}  \right) =   \cr 
  &  = \psi (2) - {1 \over 4}\psi (5/2) - {3 \over 4}\psi (3/2) =   \cr 
  &  = 1 - \gamma  - {1 \over 4}\left( {{8 \over 3} - 2\ln 2 - \gamma } \right) - {3 \over 4}\left( {2 - 2\ln 2 - \gamma } \right) =   \cr 
  &  = 2\ln 2 - {7 \over 6} \cr} 
$$
