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$\lim\limits_{n\to \infty}\left\{\large\frac{1}{1-n^2} + \frac{1}{1-n^2} +......+ \large\frac{n}{1-n^2}\right\}$ is equal to

$(a)\;0\qquad(b)\;-1/2\qquad(c)\;1/2\qquad(d)\;None\;of\;these$

1 Answer

$\lim\limits_{n\to \infty}\big(\large\frac{1}{1-n^2}+\frac{2}{1-n^2}+.....+\frac{n}{1-n^2}\big)$
$\Rightarrow \lim\limits_{n\to \infty}\large\frac{1+2+3+....+n}{1-n^2}$
$\Rightarrow \lim\limits_{n\to \infty}\large\frac{\sum n}{1-n^2}$
$\Rightarrow \lim\limits_{n\to \infty}\large\frac{n(n+1)}{2(1-n^2)}$
$\Rightarrow \lim\limits_{n\to \infty}\large\frac{1+(1/n)}{2((1/n^2)-1)}$
$\Rightarrow \large\frac{-1}{2}$
Hence (b) is the correct answer.
answered Jan 3, 2014 by sreemathi.v
 

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