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The value of $\underset{n \to \infty}{\lim} \bigg( \dfrac{1}{1-n^2} + \dfrac{2}{1-n^2} + \dots + \dfrac{n}{1-n^2} \bigg)$  is

  1. $0$
  2. $ – \frac{1}{2}$
  3. $\frac{1}{2}$
  4. none of these
in Calculus by Boss (16.8k points)
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1 Answer

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Correct Answer - Option (B)

let $L = \underset{n \to \infty}{\lim}\bigg( \frac{1}{1-n^2} + \frac{2}{1-n^2} +\frac{3}{1-n^2} + \frac{4}{1-n^2}  + ... + \frac{n}{1-n^2}\bigg) $

taking $(1 - n^2)$ common, 

$L = \underset{n \to \infty}{\lim}\bigg( \frac{1}{1-n^2} \cdot (1 + 2 + 3 + ...+n) \bigg) $

Now, sum of first $n$ natural numbers is $\frac{n(n+1)}{2}$ so substituting,

$L = \underset{x \to \infty}{\lim} \bigg( \frac{n(n+1)}{2} \cdot \frac{1}{(1-n)(1+n)}\bigg) = \frac{1}{2} \cdot \underset{x \to \infty}{\lim} \bigg( \frac{n}{1-n} \bigg)$

By L' Hopital's rule,

$\underset{x \to \infty}{\lim} \bigg( \frac{n}{1-n} \bigg) = \underset{x \to \infty}{\lim} \bigg( \frac{1}{-1} \bigg) = -1$

Hence, $L = - \frac{1}{2}$

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