Question

Solve :$\underset{n\to \infty }{lim}\left(\frac{1}{1-n^2}+\frac{2}{1-n^2}+\dots \dots \dots \dots \dots +\frac{n}{1-n^2}\right)$

• A. $1$
• B. $\frac{1}{3}$
• C. $-\frac{1}{2}$
• D. $2$

Answer 1

Answer: (C)

$-\frac{1}{2}$

Given: $\underset{n\to \infty }{lim}\left(\frac{1}{1-n^2}+\frac{2}{1-n^2}+\dots \dots \dots \dots \dots +\frac{n}{1-n^2}\right)$

$\underset{n\to \infty }{lim}\left(\frac{1}{1-n^2}+\frac{2}{1-n^2}+\dots \dots \dots \dots \dots +\frac{n}{1-n^2}\right)$

$=\underset{n\to \infty }{lim}\frac{1}{1-n^2}(1+2+3+....+n)$

$=\underset{n\to \infty }{lim}\frac{n(n+1)}{2(1-n^2)}$

$=\underset{n\to \infty }{lim}\frac{n^2(1+\frac{1}{n})}{2n^2(\frac{1}{n^2}-1)}$

$=\underset{n\to \infty }{lim}\frac{(1+\frac{1}{n})}{2(\frac{1}{n^2}-1)}$

$=-\frac{1}{2}$ $[\because n\to \infty \Rightarrow \frac{1}{n}\to 0]$

Hence option ${}^{\prime }{C}^{\prime }$ is the answer.