Question

Solve

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

• A. $0$
• B. $-1/2$
• C. $1/2$
• D. None of these

Answer 1

The correct option is B $-1/2$

Now,

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

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

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

$=\underset{n\to \infty }{lim}\frac{1+\frac{1}{n}}{2(\frac{1}{n^2}-1)}$ [ Since $\underset{n\to \infty }{lim}\frac{1}{n}=0$ and $\underset{n\to \infty }{lim}\frac{1}{n^2}=0$]

$=-\frac{1}{2}$.