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) −1/2
(c) 1/2
(d) none of these

Answer 1

(b) $\frac{-1}{2}$

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