Question

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

• A. $0$
• B. $2$
• C. $1+e^{-1}$
• D. None of the above

Answer 1

The correct option is D None of the above

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