$\lim _{n\to \infty }\left[\sum _{k=1}^{n-1}\left(1+\frac{k}{n}\right)\sin \left(\frac{k\pi }{n^2}\right)\right]$

I have to calculate this limit using Riemann sums: $$\lim _{n\to \infty }\left[\left(1+\frac{1}{n}\right)\sin \left(\frac{\pi }{n^2}\right)+\left(1+\frac{2}{n}\right)\sin \:\left(\frac{2\pi \:}{n^2}\right)+...+\left(1+\frac{n-1}{n}\right)\sin \:\left(\frac{\left(n-1\right)\pi \:}{n^2}\right)\right]$$ and $$\lim _{n\to \infty }\left[\sum _{k=1}^{n-1}\left(1+\frac{k}{n}\right)\sin \left(\frac{k\pi }{n^2}\right)\right]$$

I've done quite a few of this type of limits, but I haven't yet figured out what function $f$ to use with the Riemann sum for this particular one. Could I have some hints on how to find that function? Thank you.

Hint. Note that, for $x\geq 0$ $$x-\frac{x^3}{6}\leq \sin(x)\leq x.$$ By using the upper bound, it follows that, as $n\to\infty$, $$\sum _{k=1}^{n-1}\left(1+\frac{k}{n}\right)\sin \left(\frac{k\pi }{n^2}\right)\leq \frac{\pi}{n}\sum _{k=1}^{n-1}\left(1+\frac{k}{n}\right)\left(\frac{k }{n}\right)\to\pi\int_0^1(1+x)xdx=\frac{5\pi}{6}.$$ Now for the other side, use the lower bound, and consider that $$0\leq \sum _{k=1}^{n-1}\left(1+\frac{k}{n}\right)\left(\frac{k\pi }{n^2}\right)^3=\frac{\pi^3}{n^2}\cdot\frac{1}{n}\sum _{k=1}^{n-1}\left(1+\frac{k}{n}\right)\left(\frac{k }{n}\right)^3.$$