Question

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

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

Answer 1

Answer: (B)

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