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$\lim\limits_{n\to\infty}\bigg[\large\frac{1}{1-n^4}+\frac{1}{1-n^4}+.........+\frac{n^3}{1-n^4}\bigg]$

$(a)\;1/4\qquad(b)\;-1/4\qquad(c)\;1/2\qquad(d)\;None\;of\;these$

1 Answer

Given limit=$\lim\limits_{n\to \infty}\bigg[\large\frac{1^3+2^3+...+n^3}{1-n^4}\bigg]$
$\Rightarrow \lim\limits_{n\to \infty}\bigg[\large\frac{(n^2(n+1)^2}{4(1-n^4)}\bigg]$
$\Rightarrow \large\frac{1}{4}$$\lim\limits_{n\to \infty}\bigg(\large\frac{(1+1/n)^2}{1/n^4-1}\bigg)$
$\Rightarrow -\large\frac{1}{4}$
Hence (b) is the correct answer.
answered Dec 23, 2013 by sreemathi.v
 
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