calculus - Calculate the limit: $lim_{nrightarrowinfty}left(frac{1^{2}+2^{2}+...+n^{2}}{n^{3}}right)$

Calculate the limit: $$\lim_{n\rightarrow\infty}\left(\frac{1^{2}+2^{2}+...+n^{2}}{n^{3}}\right)$$

I'm planning to change the numerator to something else.

I know that $1+2+3+...n = \frac{n(n+1)}{2}$

And now similar just with $2$ as exponent but I did many tries on paper and always failed..

The closest I had is this but it still seems wrong:

$1^{2}+2^{2}+...+n^{2} = \frac{n(n^{2}+1)}{2}$

Well the idea is replacing numerator and then forming it, then easily calculate limit.. But I cannot find the correct thing for numerator..

Any ideas?

Answer

For variety,

$$\begin{align}
\lim_{n \to \infty} \frac{1^2 + 2^2 + \ldots + n^2}{n^3}
&=
\lim_{n \to \infty} \frac{1}{n} \left( \left(\frac{1}{n}\right)^2
+ \left(\frac{2}{n}\right)^2 + \ldots + \left(\frac{n}{n}\right)^2
\right)
\\&= \int_0^1 x^2 \mathrm{d}x = \frac{1}{3}
\end{align}
$$