$\lim_{n \rightarrow \infty}\left( \frac{1^{2}}{1 - n^{3}} + \frac{2^{2}}{1 - n^{3}} + ... + \frac{n^{2}}{1 - n^{3}} \right)$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
${\lim_{n \rightarrow \infty}\frac{1}{1 - n^{3}}}{\sum_{r = 1}^{n}{r^{2} =}\lim_{n \rightarrow \infty}\frac{1}{n^{3}\left( \frac{1}{n^{3}} - 1 \right)}\frac{n(n + 1)(2n + 1)}{6}}$
$= \lim_{n \rightarrow \infty}\frac{n^{3}\left( 1 + \frac{1}{n} \right)\left( 2 + \frac{1}{n} \right)}{n^{3}\left( \frac{1}{n^{3}} - 1 \right)6} = - \frac{1}{3}$

Q. $\lim_{n \rightarrow \infty}\left( \frac{1^{2}}{1 - n^{3}} + \frac{2^{2}}{1 - n^{3}} + ... + \frac{n^{2}}{1 - n^{3}} \right)$ is equal to

Explanation

Reviewed by Dr. Abubacker

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