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).
Using Taylor's theorem with Lagrange form of the remainder, for any $x\in \mathbb R$, $$\sin(x) = x -\frac{x^3}6 \cos(\xi_x)$$ Hence the inequality $$|\sin(x)-x|\leq \frac{|x|^3}6$$
Note next that $\displaystyle \sum_{k=1}^n \left(\frac{k\sqrt k}{n^2\sqrt n}+\frac 1{n^2}\right) = \frac 1n +\underbrace{\frac 1n \sum_{k=1}^n \frac kn \sqrt{\frac kn}}_{\text{Riemann sum}}$
and
$$\begin{align}\left| \sum_{k=1}^n \sin\left(\frac{k\sqrt k}{n^2\sqrt n}+\frac 1{n^2}\right) - \sum_{k=1}^n \left(\frac{k\sqrt k}{n^2\sqrt n}+\frac 1{n^2}\right) \right| &\leq \sum_{k=1}^n \left|\sin\left(\frac{k\sqrt k}{n^2\sqrt n}+\frac 1{n^2}\right) - \left(\frac{k\sqrt k}{n^2\sqrt n}+\frac 1{n^2}\right)\right| \\ &\leq \sum_{k=1}^n \frac 16 \left(\frac 1n + \frac 1{n^2} \right)^3 \\ &\leq \frac 16 \sum_{k=1}^n \left( \frac 2{n}\right)^3\\ &\leq \frac 43 \frac 1{n^2} \to 0 \end{align}$$
Therefore, both sum have the same limit, that is $\displaystyle \int_0^1 t\sqrt t dt$.
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.
