How do I evaluate this limit $$ \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \displaystyle \sum ^n _{k=1} \dfrac{k^2}{n^2}$$
How do I evaluate this? I have never actually learned how to work with infinite series like this one, so I have no idea.  
 A: HINT: Use the formula for the sum of the first $n$ squares:
$$\begin{align*}
\frac1n\sum_{k=1}^n\frac{k^2}{n^2}&=\frac1{n^3}\sum_{k=1}^nk^2\\
&=\frac1{n^3}\cdot\frac{n(n+1)(2n+1)}6\;.
\end{align*}$$
A: Since $n$ is a constant within the summation, you can move it outside:
$$ \displaystyle \lim_{n \to \infty} \dfrac{1}{n^3} \displaystyle \sum ^n _{k=1} {k^2}$$
The sum of the first $n$ squares is well known to be $\dfrac{n^3}{3} +\dfrac{n^2}{2}+\dfrac{n}{6}$. 
Dividing by $n^3$, we have $\dfrac{1}{3} +\dfrac{1}{2n}+\dfrac{1}{n^2}$. The limit as $n \to \infty$ is now obvious as $\dfrac{1}{3}$.
A: 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
$$\lim_{\infty}\frac{1}{n^3}\sum_{k=1}^n k^2=\frac{1}{3}.$$
A: You can recognize a Riemann sum: $$\lim\limits_{n \to + \infty} \frac{1}{n} \sum\limits_{k=1}^n \frac{k^2}{n^2} = \int_0^1 x^2dx= \frac{1}{3}$$
