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.
 A: By induction, you can prove that
$\displaystyle \sum_{k = 1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}$
Thus
$\displaystyle \lim_{n \rightarrow \infty}\frac{1}{n^3}\sum_{k = 1}^{n - 1} k^2 = \lim_{n \rightarrow \infty}\frac{1}{n^3}\frac{(n - 1)(n)(2(n - 1) + 1)}{6} = \lim_{n \rightarrow \infty}\frac{1}{n^3}\frac{(n - 1)(n)(2n - 1)}{6} = \lim_{n \rightarrow \infty}\frac{2n^3 - 3n^2 + n}{6n^3}$
$\displaystyle \lim_{n \rightarrow \infty}\frac{1}{3} - \frac{1}{2n} + \frac{1}{6n^2} = \frac{1}{3}$
A: Well done. The Riemann sum argument was carried out efficiently. 
For another way of solving the problem, we can use the fact that the sum $\sum_{k=1}^n k^2$ is given by the explicit formula 
$$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}.$$
I have quoted the sum to $n$, because it is the standard version, but you can get the explicit formula for the sum to $n-1$ by making the obvious substitution.
There are many ways to prove the above formula. One way is by induction. 
So (after the substitution) we want to find the limit of 
$$\frac{(n-1)(n)(2n-1)}{6n^3}.$$
Cancel an $n$, then divide top and bottom by $n^2$. We want
$$\lim_{n\to\infty}\frac{\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)}{6}.$$
Remark: For additional explicit "sums of powers" formulas, please see the Wikipedia article on Faulhaber's Formula.
A: What you have is fine. If one knows the formula $\sum_{k=1}^nk^2=\frac16n(n+1)(2n+1)$, one can avoid the integration:
$$\sum_{k=1}^{n-1}k^2=\frac16n(n-1)(2n-1)=\frac{2n^3-3n^2+n}6\;,$$
so $$\lim_{n\rightarrow \infty} \frac{1}{n^{3}}\sum_{k=1}^{n-1}k^2=\lim_{n\to\infty}\frac{2-\frac3n+\frac1n}6=\frac13\;.$$
A: 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
$\displaystyle \lim_{n\rightarrow \infty} \frac{1}{n^{3}}\sum_{k=1}^{n-1}k^2 =\frac{1}{3}$
A: Also, if you know it, this is a straightforward application of Stoltz-Cezaro:
$$\lim_{n\rightarrow \infty} \frac{1}{n^{3}}\sum_{k=1}^{n-1}k^2=\lim_{n\rightarrow \infty} \frac{\sum_{k=1}^{n}k^2-\sum_{k=1}^{n-1}k^2}{(n+1)^{3}-n^3}= \lim_{n\rightarrow \infty} \frac{n^2}{3n^2+3n+1}=\frac{1}{3}$$
