Respuesta :
[tex]\\ \sf\Rrightarrow \lim_{n\to \infty}\left(\dfrac{1}{1-n^2}+\dfrac{2}{2-n^2}\dots \dfrac{n}{1-n^2}\right)[/tex]
- Take LCM as 1-n^2
[tex]\\ \sf\Rrightarrow \lim_{n\to \infty}\left(\dfrac{1+2+3\dots n}{1-n^2}\right)[/tex]
- 1+2..n=n(n+1)/2
[tex]\\ \sf\Rrightarrow \lim_{n\to \infty}\left(\dfrac{\dfrac{n(n+1)}{2}}{1-n^2}\right)[/tex]
[tex]\\ \sf\Rrightarrow \lim_{n\to \infty}\dfrac{n(n+1)}{2(1-n^2)}[/tex]
[tex]\\ \sf\Rrightarrow \lim_{n\to infty}\dfrac{n(1+n)}{2(1-n)(1+n)}[/tex]
[tex]\\ \sf\Rrightarrow \lim_{n\to \infty}\dfrac{n}{2(1-n)}[/tex]
[tex]\\ \sf\Rrightarrow \dfrac{\infty}{2-\infty}[/tex]
[tex]\\ \sf\Rrightarrow \dfrac{-1}{2}[/tex]
[tex]{\sf \lim_{n \to \infty} (\frac{1}{1-n^2}+\frac{2}{1-n^2} +...\: \frac{n}{1-n^2}) } \\ = {\sf \lim_{n \to \infty} (\frac{1 + 2 + ..n}{1-n^2})} \\ = {\sf \lim_{n \to \infty} ( \frac{n(n + 1)}{2} \div 1 - {n}^{2} )} \\ = {\sf \lim_{n \to \infty} ( \frac{n(n + 1)}{2} \times \frac{1}{ 1 - {n}^{2}} )} \\ ={\sf \lim_{n \to \infty} ( \frac{n(n + 1)}{2(1 - {n}^{2} )} )} \\ = {\sf \lim_{n \to \infty} ( \frac{n(n + 1)}{2 (1 - n )(1 + n) } )} \\ = {\sf \lim_{n \to \infty} ( \frac{n(1 + n)}{2 (1 - n )(1 + n) } )} \\ = {\sf \lim_{n \to \infty} ( \frac{n}{2 (1 - n )} )} \\ = {\sf \lim_{n \to \infty} ( \frac{n}{2 - 2 n } )} \\ = \sf \frac{ \infty }{2 - \infty } \\ = \frac{ - 1}{2}
[/tex]
Answer:
[tex] \frac{ - 1}{2} [/tex]
Hope you could get an idea from here.
Doubt clarification - use comment section.
