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$\lim\limits_{n\to \infty}\bigg[\large\frac{1}{1-n^2}+\frac{2}{1-n^2}+..........+\frac{n}{1-n^2}\bigg]$ is

$(a)\;0\qquad(b)\;-\large\frac{1}{2}\qquad$$(c)\;\large\frac{1}{2}$$\qquad(d)\;None\;of\;these$

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1 Answer

$\lim\limits_{n\to \infty}\bigg[\large\frac{1}{1-n^2}+\frac{2}{1-n^2}+.....+\frac{n}{1-n^2}\bigg]$
$\Rightarrow \lim\limits_{n\to \infty}\bigg[\large\frac{1+2+.....+n}{1-n^2}\bigg]=$$\lim\limits_{n\to \infty}\large\frac{n(n+1)}{2(1-n^2)}$
$\Rightarrow \lim\limits_{n\to \infty}\large\frac{1+1/n}{2(\Large\frac{1}{n^2}-1)}$
$\Rightarrow \large\frac{1+0}{2(0-1)}$
$\Rightarrow -\large\frac{1}{2}$
Hence (b) is the correct answer.
answered Jan 2, 2014 by sreemathi.v
 

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