Question

${lim}_{n\to \infty }[\frac{1}{1-n^2}+\frac{2}{1-n^2}+\cdots +\frac{n}{1-n^2}]$ is equal to

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

Answer 1

The correct option is B

$\frac{-1}{2}$

${lim}_{n\to \infty }[\frac{1}{1-n^2}+\frac{2}{1-n^2}+\cdots +\frac{n}{1-n^2}]$
$={lim}_{n\to \infty }\left(\frac{1}{1-n^2}\right)[1+2+3\cdots n]$
$={lim}_{n\to \infty }\left(\frac{1}{1-n^2}\right)\left[\frac{n(n+1)}{2}\right]$
$={lim}_{n\to \infty }\frac{1}{2}\left[\frac{n(n+1)}{(1-n)(1+n)}\right]$
$={lim}_{n\to \infty }\frac{1}{2}\frac{n}{1-n}$
$={lim}_{n\to \infty }\frac{1}{2}\frac{1}{\frac{1}{n}-1}$
$=\frac{1}{2}(-1)$
$=\frac{-1}{2}$