How to calculate limit $S_n=\sum\limits_{k=1}^n\sin\Big( \dfrac{k\sqrt{k}}{n^2\sqrt{n}}+\dfrac{1}{n^2}\Big)$ 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 $\displaystyle \int_0^1 t\sqrt t dt$.
