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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$\frac{5}{16}$

Vedclass · Answer

(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

Similar Questions

If $\alpha, \beta$ are the roots of the equation $x^2 - 3x + a = 0$ and $\gamma, \delta$ are the roots of the equation $x^2 - 12x + b = 0$,and $\alpha, \beta, \gamma, \delta$ form an increasing $G.P.$,then $(a, b) = $

The maximum sum of the series $20 + 19\frac{1}{3} + 18\frac{2}{3} + \dots$ is

The first term of an $A.P.$ is $2$ and the common difference is $4$. The sum of its $40$ terms will be:

If $a, b, c$ are the $p^{th}, q^{th}$ and $r^{th}$ terms of a $G.P.$,then $\left( \frac{c}{b} \right)^p \left( \frac{b}{a} \right)^r \left( \frac{a}{c} \right)^q$ is equal to

The sum to $n$ terms of the series $1^2 + (1^2 + 3^2) + (1^2 + 3^2 + 5^2) + ...$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module