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Question: \(\lim_{n \rightarrow \infty}\left\lbrack \frac{1}{1 - n^{2}} + \frac{2}{1 - n^{2}} + ...... + \frac...

lim⁡n→∞[11−n2+21−n2+......+n1−n2]\lim_{n \rightarrow \infty}\left\lbrack \frac{1}{1 - n^{2}} + \frac{2}{1 - n^{2}} + ...... + \frac{n}{1 - n^{2}} \right\rbrack is equal to

A

0

B

−12- \frac{1}{2}

C

12\frac{1}{2}

D

None of these

Answer

−12- \frac{1}{2}

Explanation

Solution

lim⁡n→∞[11−n2+21−n2+......+n1−n2]\lim_{n \rightarrow \infty}\left\lbrack \frac{1}{1 - n^{2}} + \frac{2}{1 - n^{2}} + ...... + \frac{n}{1 - n^{2}} \right\rbrack

=lim⁡n→∞∑n1−n2=12lim⁡n→∞n2+n1−n2=−12.= \lim_{n \rightarrow \infty}\frac{\sum n}{1 - n^{2}} = \frac{1}{2}\lim_{n \rightarrow \infty}\frac{n^{2} + n}{1 - n^{2}} = - \frac{1}{2}.