TITLE: How to calculate limit $S_n=\sum\limits_{k=1}^n\sin\Big( \dfrac{k\sqrt{k}}{n^2\sqrt{n}}+\dfrac{1}{n^2}\Big)$
QUESTION [5 upvotes]: How do I calculate the following limit:
 $$\lim\limits_{n \to +\infty} S_n=\lim\limits_{n \to +\infty}\sum\limits_{k=1}^n\sin\Big( \dfrac{k\sqrt{k}}{n^2\sqrt{n}}+\dfrac{1}{n^2}\Big) = \text{?}$$
I think that you need to use Riemann sum, but I don't understand how to get rid of the sine.
Please provide a hint (and not the full solution).

REPLY [2 votes]: Generalization: Suppose $f$ is any function on $[0,2]$ with $f(0)=0$ such that $f'(0)$ exists. Then
$$\tag 1 \lim_{n\to\infty} \sum_{k=1}^{n}f\left(\frac{k^{3/2}}{n^{5/2}} + \frac{1}{n^2}\right ) = \frac{2f'(0)}{5}.$$
Brief sketch: $f(x) = f'(0)x + o(x)$ as $x\to 0.$ It follows that after simlifying, the sum in $(1)$ equals
$$\sum_{k=1}^{n}\left [f'(0)\left (\frac{k^{3/2}}{n^{5/2}} + \frac{1}{n^2}\right ) + o(1/n)\right] = f'(0)\left (\sum_{k=1}^{n}\frac{k^{3/2}}{n^5/2}\right ) + o(1).$$
In the usual way you see the limit of the last sum is $\int_0^1 t^{3/2}\,dt = 2/5.$ The claimed result follows.