Question

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

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

Answer 1

The correct option is B

$-\frac{1}{2}$

Explanation for the correct option:

Finding the value of the given limit:

Given that,

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

Applying the limits,

$$\begin{gathered} =\left(\frac{1+{\displaystyle \frac{1}{\infty }}}{2\left({\displaystyle \frac{1}{\infty }}-1\right)}\right) \\ =\left(\frac{1+0}{2(0-1)}\right) \\ =-\frac{1}{2} \end{gathered}$$

Therefore, the correct answer is option (B).