Evaluate _ n 1^2+2^2+3^2+ +n^2 n^3 .

Question

Evaluate $\lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots\ldots+n^2}{n^3}.$

✓

Answer

We know that $1^2+2^2+3^2+\ldots \ldots \ldots+n^2=\Sigma n^2 =\frac{1}{6} n(n+1)(2 n+1)$
$\therefore \lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots \ldots+n^2}{n^3}$
$=\lim _{n \rightarrow \infty} \frac{\frac{1}{6} n(n+1)(2 n+1)}{n^3}=\frac{1}{6} \times \lim _{n \rightarrow \infty}\left(\frac{n+1}{n}\right)\left(\frac{2 n+1}{n}\right)$
$=\frac{1}{6} \times \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)=\frac{1}{6}\left(1+\frac{1}{\infty}\right)\left(2+\frac{1}{\infty}\right)$
$=\frac{1}{6} \times(1+0)(2+0)=\frac{1}{6} \times 2=\frac{1}{3}$

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