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Question: \(\lim _ { n \rightarrow \infty }\) \(\left( \frac{1}{1 - n^{2}} + \frac{2}{1 - n^{2}} + ...... + \f...

lim⁡n→∞\lim _ { n \rightarrow \infty } (11−n2+21−n2+......+n1−n2)\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

½

D

None of these

Answer

–1/2

Explanation

Solution

(11−n2+21−n2+......+n1−n2)\left( \frac{1}{1 - n^{2}} + \frac{2}{1 - n^{2}} + ...... + \frac{n}{1 - n^{2}} \right)

=11−n2\frac{1}{1 - n^{2}}× (1 + 2 + ..... + n)

= 11−n2\frac{1}{1 - n^{2}}. n(n+1)2\frac{n(n + 1)}{2}= n2(1−n)\frac{n}{2(1 - n)}= 12[(1/n)−1]\frac{1}{2\lbrack(1/n) - 1\rbrack}

∴  lim⁡n→∞\lim _ { n \rightarrow \infty } (11−n2+21−n2+......+n1−n2)\left( \frac{1}{1 - n^{2}} + \frac{2}{1 - n^{2}} + ...... + \frac{n}{1 - n^{2}} \right)

= lim⁡n→∞\lim _ { n \rightarrow \infty } 12[(1/n)−1]\frac{1}{2\lbrack(1/n) - 1\rbrack}= – 12\frac{1}{2}