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 $y = x - x^2 + x^3 - x^4 + \dots \infty$,then the value of $x$ will be

Which term of the geometric sequence $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683}$?

Find the sum of $1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots$ to infinite terms.

The sum of the first five terms of the series $3 + 4\frac{1}{2} + 6\frac{3}{4} + \dots$ will be

If the $p^{th}$ term of an $A.P.$ is $q$ and the $q^{th}$ term is $p,$ then its $r^{th}$ term is

Vedclass Test Series

Exam Paper Generator

Online Exam Module