Find the limits of Riemann Sum

Question

Find the $\lim_{n \to \infty} \frac {1}{n^3} (1^2 \cos \frac {2}{n} + 2^2 \cos \frac {4}{n} + \cdots + n^2 \cos \frac {2n}{n})$

This seems to be a Riemann Sum, but the question lies at how should I deduce the interval?

I've got $f(x) = x^2 \cos (2x)$ (which i'm not sure if it's right?), $Δx = 1/n$, $x_{i} = i/n$, if i let the interval to be $[0, 1]$.

Does that mean $$=\frac{1}{n}\sum_{i=1}^n (i/n)^2 \cos (2i/n)$$

Probably somewhere is wrong, and I don't get the whole conceptual ideas, if it's possible please give some explanations as well thanks.

Answer 1

Your wrong is that you should use $f(x_i)$, i.e. the sum is of the form $$\frac{1}{n}\sum_{i=1}^n\frac{i^2}{n^2}\cos(\frac{2i}{n}). $$

Answer 2

You wrote the argument of the limit in the wrong way. It is $$ \frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^2 \cos\left(\frac{2k}{n}\right) $$ and the limit of such object as $n\to +\infty$ is $$ \int_{0}^{1} x^2\cos(2x)\,dx \stackrel{\text{IBP}}{=} \frac{\cos 2}{2}+\frac{\sin 2}{4}$$ by the definition of Riemann-integrability.