How do I evaluate this limit

Question

$$ \displaystyle \lim_{n \to \infty} \dfrac{1}{n} \displaystyle \sum ^n _{k=1} \dfrac{k^2}{n^2}$$

How do I evaluate this? I have never actually learned how to work with infinite series like this one, so I have no idea.

Answer 1

You can recognize a Riemann sum: $$\lim\limits_{n \to + \infty} \frac{1}{n} \sum\limits_{k=1}^n \frac{k^2}{n^2} = \int_0^1 x^2dx= \frac{1}{3}$$

Answer 2

HINT: Use the formula for the sum of the first $n$ squares:

$$\begin{align*} \frac1n\sum_{k=1}^n\frac{k^2}{n^2}&=\frac1{n^3}\sum_{k=1}^nk^2\\ &=\frac1{n^3}\cdot\frac{n(n+1)(2n+1)}6\;. \end{align*}$$

Answer 3

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 $\dfrac{1}{3}$.

Answer 4

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 $$\lim_{\infty}\frac{1}{n^3}\sum_{k=1}^n k^2=\frac{1}{3}.$$