The value of sumlimits_{n = 2}^infty {frac{n | vedclass

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Question

(C) The given expression is $\sum\limits_{n = 2}^\infty  {\frac{n}{{1 + {n^4} - 2{n^2}}}} = \sum\limits_{n = 2}^\infty  {\frac{n}{{{{(n^2 - 1)}^2}}}}

Vedclass · Accepted Answer

$\frac{5}{16}$

Vedclass · Answer

(C) The given expression is $\sum\limits_{n = 2}^\infty  {\frac{n}{{1 + {n^4} - 2{n^2}}}} = \sum\limits_{n = 2}^\infty  {\frac{n}{{{{(n^2 - 1)}^2}}}} $.We can rewrite the term as $\frac{n}{{{{(n-1)}^2}{(n+1)}^2}}$.Using partial fractions,$\frac{n}{{{{(n-1)}^2}{(n+1)}^2}} = \frac{1}{4} \left( \frac{1}{{{{(n-1)}^2}}} - \frac{1}{{{{(n+1)}^2}}} \right)$.Now,the sum becomes $\frac{1}{4} \sum\limits_{n = 2}^\infty  {\left( {\frac{1}{{{{(n - 1)}^2}}} - \frac{1}{{{{(n + 1)}^2}}}} \right)}$.This is a telescoping series:$= \frac{1}{4} \left[ \left( \frac{1}{{{1^2}}} - \frac{1}{{{3^2}}} \right) + \left( \frac{1}{{{2^2}}} - \frac{1}{{{4^2}}} \right) + \left( \frac{1}{{{3^2}}} - \frac{1}{{{5^2}}} \right) + \dots \right]$.$= \frac{1}{4} \left( \frac{1}{1} + \frac{1}{4} \right) = \frac{1}{4} \left( \frac{5}{4} \right) = \frac{5}{16}$.

The value of $\sum\limits_{n = 2}^\infty {\frac{n}{{1 + {n^2}\left( {{n^2} - 2} \right)}}} $ is equal to

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