A rigorous proof for a Riemann-like sum

For any $x\in[0,1]$ we have $x-\frac{x^3}{6}\leq\sin(x)\leq x$. We have that $$ \lim_{n\to +\infty}\sum_{k=1}^{n}\sin\left(\frac{k}{n}\right)\frac{k}{n^2} = \int_{0}^{1}x\sin(x)\,dx = \sin(1)-\cos(1)\tag{1}$$ since $x\sin(x)$ is a continuous function on $[0,1]$, hence a Riemann-integrable function.
For the same reason $$ \lim_{n\to +\infty}\sum_{k=1}^{n}\sin\left(\frac{k}{n}\right)\frac{k^3}{n^4} = \int_{0}^{1}x^3\sin(x)\,dx = 5\cos(1)-3\sin(1)\tag{2}$$ hence $\sum_{k=1}^{n}\sin\left(\frac{k}{n}\right)\frac{k^3}{6n^6}=O\left(\frac{1}{n^2}\right)$ and we are allowed to simply replace $\sin\left(\frac{k}{n^2}\right)$ with $\frac{k}{n^2}$ in the given sum: the difference of the associated sums is negligible for large values of $n$.

This is similar to Jack D'Aurizio's answer, but I think it might be clearer to use big-O notation. Here we can use that $\sin(x)=x+O\!\left(x^3\right)$: $$ \begin{align} \lim_{n\to\infty}\sum_{k=1}^n\sin\left(\frac{k}{n}\right)\sin\left(\frac{k}{n^2}\right) &=\lim_{n\to\infty}\sum_{k=1}^n\sin\left(\frac{k}{n}\right)\left(\frac{k}{n^2}+O\left(\frac{k^3}{n^6}\right)\right)\\ &=\lim_{n\to\infty}\sum_{k=1}^n\sin\left(\frac{k}{n}\right)\frac{k}{n}\frac1n+\lim_{n\to\infty}O\left(\frac1{n^2}\sum_{k=1}^n\sin\left(\frac{k}{n}\right)\frac{k^3}{n^3}\frac1n\right)\\ &=\int_0^1\sin(x)\,x\,\mathrm{d}x+\lim_{n\to\infty}O\!\left(\frac1{n^2}\right)\int_0^1\sin(x)\,x^3\,\mathrm{d}x\\[3pt] &=\int_0^1\sin(x)\,x\,\mathrm{d}x\\[3pt] &=-\cos(1)+\int_0^1\cos(x)\,\mathrm{d}x\\[9pt] &=\sin(1)-\cos(1) \end{align} $$