A) 0
B) \[-\frac{1}{2}\]
C) \[\frac{1}{2}\]
D) None of these
Correct Answer: B
Solution :
\[\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{1}{1-{{n}^{2}}}+\frac{2}{1-{{n}^{2}}}+...+\frac{n}{1-{{n}^{2}}} \right)\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{1+2+...+n}{1-{{n}^{2}}} \right)\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n(n+1)}{2(1-{{n}^{2}})}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\left( 1+\frac{1}{n} \right)}{2\left( \frac{1}{{{n}^{2}}}-1 \right)}=-\frac{1}{2}\]
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