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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
recategorized | 11 views

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,

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

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

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

by Active (1.8k points)