Question

$\underset{x\to \infty }{lim}(\frac{1}{1-n^2}+\frac{2}{1-n^2}+\frac{3}{1-n^2}+...+\frac{n}{1-n^2})$

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is D $-\frac{1}{2}$

Consider the given function,

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

$\Rightarrow \underset{n\to \infty }{lim}\frac{1}{1-n^2}(1+2+3+\mathrm{4........}n)$

$\Rightarrow \underset{n\to \infty }{lim}\frac{1}{1-n^2}\frac{n(n+1)}{2}$

$\Rightarrow \underset{n\to \infty }{lim}\frac{1}{(1-n)(1+n)}\frac{n(n+1)}{2}$

$\Rightarrow \frac{1}{2}\underset{n\to \infty }{lim}\frac{n}{(1-n)}$

Apply L-Hospital rule,

$\Rightarrow \frac{1}{2}\underset{n\to \infty }{lim}\frac{1}{-1}$

$\Rightarrow \frac{1}{2}\times (-1)$

$\Rightarrow -\frac{1}{2}$

Hence, this is the answer.