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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.

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  • $\begingroup$ You divided your convergent series into three divergent ones. Doesn't look very promising...unless you can produce a telescopic series of something like that (hint) $\endgroup$ – DonAntonio Jan 24 at 23:01
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Hint: $\frac {n+1} {n(2n-1)(2n+1)}=\frac 3 4 \frac 1 {n(2n-1)} -\frac 1 4 \frac 1 {n(2n+1)}$. So what is left is to find the sums $\sum \frac 1 {n(2n-1)}$ and $\sum \frac 1 {n(2n+1)}$.

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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 $$ \eqalign{ & S = \mathop {\lim }\limits_{N\, \to \,\infty } \left( \matrix{ {1 \over 4}\psi (N + 5/2) + {3 \over 4}\psi (N + 3/2) - \psi (N + 2) + \hfill \cr + \psi (2) - {1 \over 4}\psi (5/2) - {3 \over 4}\psi (3/2) \hfill \cr} \right) = \cr & = \psi (2) - {1 \over 4}\psi (5/2) - {3 \over 4}\psi (3/2) = \cr & = 1 - \gamma - {1 \over 4}\left( {{8 \over 3} - 2\ln 2 - \gamma } \right) - {3 \over 4}\left( {2 - 2\ln 2 - \gamma } \right) = \cr & = 2\ln 2 - {7 \over 6} \cr} $$

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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 $$\sum_{n = 2}^\infty \frac{n + 1}{n(2n - 1)(2n + 1)} = 2 \ln 2 - \frac{7}{6}.$$

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  • $\begingroup$ another approach to confirm the correctness of my answer: thanks. $\endgroup$ – G Cab Jan 26 at 19:26

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