$\lim_{n \rightarrow \infty}\frac{1^{2} + 2^{2} + 3^{2} + ... + n^{2}}{n^{3}}$ is equal to

Question

Dr. Abubacker · Accepted Answer

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

Q. $\lim_{n \rightarrow \infty}\frac{1^{2} + 2^{2} + 3^{2} + ... + n^{2}}{n^{3}}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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