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

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(C) The given expression is $S = \sum_{n=2}^{\infty} \frac{n}{1 + n^2(n^2 - 2)} = \sum_{n=2}^{\infty} \frac{n}{n^4 - 2n^2 + 1}$.Note that the denomina

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$\frac{5}{16}$

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(C) The given expression is $S = \sum_{n=2}^{\infty} \frac{n}{1 + n^2(n^2 - 2)} = \sum_{n=2}^{\infty} \frac{n}{n^4 - 2n^2 + 1}$.Note that the denominator is a perfect square: $n^4 - 2n^2 + 1 = (n^2 - 1)^2 = ((n-1)(n+1))^2 = (n-1)^2(n+1)^2$.Thus,the term is $\frac{n}{(n-1)^2(n+1)^2}$.Using partial fractions,we can write $\frac{n}{(n-1)^2(n+1)^2} = \frac{1}{4} \left( \frac{1}{(n-1)^2} - \frac{1}{(n+1)^2} \right)$.Now,we evaluate the telescoping sum:$S = \frac{1}{4} \sum_{n=2}^{\infty} \left( \frac{1}{(n-1)^2} - \frac{1}{(n+1)^2} \right)$.Expanding the sum:$S = \frac{1}{4} \left[ \left( \frac{1}{1^2} - \frac{1}{3^2} \right) + \left( \frac{1}{2^2} - \frac{1}{4^2} \right) + \left( \frac{1}{3^2} - \frac{1}{5^2} \right) + \left( \frac{1}{4^2} - \frac{1}{6^2} \right) + \dots \right]$.Most terms cancel out,leaving only the first two positive terms:$S = \frac{1}{4} \left( 1 + \frac{1}{4} \right) = \frac{1}{4} \left( \frac{5}{4} \right) = \frac{5}{16}$.

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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