Compute this limit of series: $\lim_{n\rightarrow \infty} \frac{1}{n^{3}}\sum_{k=1}^{n-1}k^2 $ Compute this limit of series:
$\displaystyle \lim_{n\rightarrow \infty} \frac{1}{n^{3}}\sum_{k=1}^{n-1}k^2 $
I used the definition of the definite integral
$\displaystyle \int_{a}^{b}f(x)dx=\lim_{n\rightarrow \infty} S_{n}$
where $\displaystyle f(x)=x^2$, $\displaystyle [a,b]=[0,1]$; since the function is continuous in $\displaystyle [0, 1]$ then it is certainly integrated.
$\displaystyle \int_{0}^{1}x^2dx=\lim_{n\rightarrow \infty} \sum_{k=1}^{n-1}\frac{1}{n}\ \left(\frac{k}{n}\right)^2=\lim_{n\rightarrow \infty} \frac{1}{n^3}\sum_{k=1}^{n-1}k^2$
We have:
$\displaystyle\int_{0}^{1}x^2dx=\frac{1}{3}$
then
$\displaystyle \lim_{n\rightarrow \infty} \frac{1}{n^{3}}\sum_{k=1}^{n-1}k^2 =\frac{1}{3}$
Any suggestions, please? This limit can be solved in other ways?
Thanks.

By induction, you can prove that
$\displaystyle \sum_{k = 1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}$
Thus
$\displaystyle \lim_{n \rightarrow \infty}\frac{1}{n^3}\sum_{k = 1}^{n - 1} k^2 = \lim_{n \rightarrow \infty}\frac{1}{n^3}\frac{(n - 1)(n)(2(n - 1) + 1)}{6} = \lim_{n \rightarrow \infty}\frac{1}{n^3}\frac{(n - 1)(n)(2n - 1)}{6} = \lim_{n \rightarrow \infty}\frac{2n^3 - 3n^2 + n}{6n^3}$
$\displaystyle \lim_{n \rightarrow \infty}\frac{1}{3} - \frac{1}{2n} + \frac{1}{6n^2} = \frac{1}{3}$
