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.