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Mathematics Question on limits and derivatives

\displaystyle \lim_{n \to\infty} \left\\{ \frac{1}{1-n^{2}} + \frac{2}{1-n^{2}} + .... + \frac{n}{1-n^{2}}\right\\} 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)\displaystyle \lim_{n \to\infty} \left(\frac{1}{1-n^{2}} + \frac{2}{1-n^{2}} + .... + \frac{n}{1-n^{2}}\right)
=lim⁡n→∞1+2+3+....+n1−n2=\displaystyle \lim_{n \to\infty} \frac{ 1+2+3+....+n}{1-n^{2}}
=lim⁡n→∞∑n1−n2= \displaystyle \lim_{n\to\infty} \frac{\sum n}{1-n^{2}}
=lim⁡n→∞n(n+1)2(1−n2)= \displaystyle \lim_{n \to\infty} \frac{n\left(n+1\right)}{2\left(1-n^{2}\right)}
=lim⁡n→∞1+1/n2[1n2−1]= \displaystyle \lim_{n\to\infty} \frac{1+1/n}{2\left[ \frac{1}{n^{2} } - 1 \right]}
=−1/2= - 1/2