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.

Here is a way to find those two sums @Kavi Rama Murthy refers to in his answer. They are found by first converting them to double integrals.
Let
$$S = \sum_{n = 2}^\infty \frac{n + 1}{n(2n - 1)(2n + 1)} = \frac{3}{4} \sum_{n = 2}^\infty \frac{1}{n (2n - 1)} - \frac{1}{4} \sum_{n = 1}^\infty \frac{1}{n(2n + 1)} + \frac{1}{12}. \qquad (*)$$
For the first of the sums appearing to the right in ($*$), observe that
$$\frac{1}{n} = \int_0^1 x^{n - 1} \, dx \quad \text{and} \quad \frac{1}{2n - 1} = \int_0^1 y^{2n - 2} \, dy.$$
The sum may now be rewritten as
\begin{align}
\sum_{n = 2}^\infty \frac{1}{n(2n - 1)} &= \sum_{n = 2}^\infty \int_0^1 \int_0^1 x^{n - 1} y^{2n - 2} \, dx \, dy\\
&= \int_0^1 \int_0^1 \frac{1}{xy^2} \sum_{n = 2}^\infty (xy^2)^n \, dx \,dy \tag1\\
&= \int_0^1 \int_0^1 \frac{1}{xy^2} \cdot \frac{(xy^2)^2}{1 - xy^2} \, dx \, dy \tag2\\
&= \int_0^1 \frac{1}{y^2} \int_0^{y^2} \frac{u}{1 - u} \, du \, dy \tag3\\
&= -\int_0^1 \frac{1}{y^2} \int_0^{y^2} \left (1 - \frac{1}{1 - u} \right ) \, du \, dy\\
&= -\int_0^1 \left (1 + \frac{\ln (1 - y^2)}{y^2} \right ) \, dy\\
&= -1 + 2 \int_0^1 \frac{1 - y}{1 - y^2} \, dy \tag4\\
&= -1 + 2 \int_0^1 \frac{dy}{1 + y}\\
&= -1 + 2 \ln 2.
\end{align}
Explanation
(1) Interchange of the sum and integral signs is justified by the dominated convergence theorem.
(2) Summing of a geometric series.
(3) Change of variable $u = xy^2$ is made.
(4) Integration by parts has been used.
The second of the sums appearing in ($*$) can be found in a similar manner (see here). The result is:
$$\sum_{n = 1}^\infty \frac{1}{n(2n + 1)} = 2 - 2\ln 2.$$
Thus
$$S = \frac{3}{4} (2 \ln 2 - 1) - \frac{1}{4} (2 - 2 \ln 2) + \frac{1}{12},$$
or
$$\sum_{n = 2}^\infty \frac{n + 1}{n(2n - 1)(2n + 1)} = 2 \ln 2 - \frac{7}{6}.$$
