Will the expression $\sum_{i=1}^{n}{\frac{i^{2}}{n^{2}}}$ converge as n approches infinity?

Question

I have the following expression:

$$\lim_{n \to\infty}\ \sum_{i=1}^{n}{(\frac{i}{n})^{2}}$$

I am not quite sure whether it will converge or diverge. Can somebody tell me how to figure it out?

Answer 1

$$\begin{align} & \lim_{n \to\infty}\ \sum_{i=1}^{n}{(\frac{i}{n})^{2}} \\ & =\lim_{n \to\infty}\ \frac{1}{n^{2}}\sum_{i=1}^{n}i^{2}\\ & =\lim_{n \to\infty}\ \frac{1}{n^{2}}\cdot \frac{n(n+1)(2n+1)}{6}\\ & =\lim_{n \to\infty}\ \frac{1}{6}\cdot (1+\frac{1}{n})(2n+1)\\ & =\frac{1}{6}\cdot (1+ 0)\cdot \infty\\ & = \infty \end{align}$$

The sequence clearly diverges.

Answer 2

No. It diverges.

$ \sum_{i=1}^{n}{(\frac{i}{n})^{2}} =\frac1{n^2}\sum_{i=1}^{n} i^2 $ and $\sum_{i=1}^{n} i^2 =\frac16 n(n+1)(2n+1) \approx \frac13 n^3 $, so $ \sum_{i=1}^{n}{(\frac{i}{n})^{2}} \approx \frac13 n $.

Answer 3

Maybe you can consider $$\lim_{n\to \infty}\frac1n\sum_{i=1}^n\left(\frac {i}{n}\right)^2=\int_0^1x^2\mathrm dx=\frac13$$ Then easy to get the original limit is $\infty$.