TITLE: How do I evaluate this limit
QUESTION [5 upvotes]: $$ \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \displaystyle \sum ^n _{k=1} \dfrac{k^2}{n^2}$$
How do I evaluate this? I have never actually learned how to work with infinite series like this one, so I have no idea.

REPLY [2 votes]: Let's remember a usefull result on the series: if the serie $\displaystyle \sum_n f(n)$ is divergent then we have
$$\sum_{k=1}^n f(k)\sim\int_1^nf(x)dx.$$
With this result we find the classical result
$$\sum_{k=1}^n\frac{1}{n}\sim\int_1^n\frac{dx}{x}=\log n.$$
Now, we return to the question. Since the serie $\displaystyle \sum_n n^2$ is divergent then we have
$$\sum_{k=1}^n k^2\sim \int_1^n x^2dx=\frac{1}{3}\left[x^3\right]_1^n\sim\frac{n^3}{3},$$
and hence we find
$$\lim_{\infty}\frac{1}{n^3}\sum_{k=1}^n k^2=\frac{1}{3}.$$