Question

$ \displaystyle \lim_{n \to\infty} \left\{ \frac{1}{1-n^{2}} + \frac{2}{1-n^{2}} + .... + \frac{n}{1-n^{2}}\right\}$ is equal to

• A. 0
• B. $ - \frac{1}{2}$
• C. $ \frac{1}{2}$
• D. none of these

Answer 1

The Correct Option is B

$\displaystyle \lim_{n \to\infty} \left(\frac{1}{1-n^{2}} + \frac{2}{1-n^{2}} + .... + \frac{n}{1-n^{2}}\right)$
$=\displaystyle \lim_{n \to\infty} \frac{ 1+2+3+....+n}{1-n^{2}} $
$= \displaystyle \lim_{n\to\infty} \frac{\sum n}{1-n^{2}}$
$ = \displaystyle \lim_{n \to\infty} \frac{n\left(n+1\right)}{2\left(1-n^{2}\right)} $
$= \displaystyle \lim_{n\to\infty} \frac{1+1/n}{2\left[ \frac{1}{n^{2} } - 1 \right]}$
$ = - 1/2$