Compute $$\lim_{n\rightarrow\infty}\sum_{i=1}^n\sin\left(\frac{i}{n^2}\right)$$
Using Taylor expansion for $\sin x$, I know that this is $$\lim_{n\rightarrow\infty}\sum_{i=1}^n\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)!}\left(\frac{i}{n^2}\right)^{2k+1}$$ How do I transform this into a Riemann integral?
Let $S_n=\sum\limits_{i=1}^n\sin(i/n^2)$. For every $x\geqslant0$, $x-x^3\leqslant\sin x\leqslant x$ hence $$ \frac1{n^2}T_n-\frac1{n^6}R_n\leqslant S_n\leqslant \frac1{n^2}T_n, \qquad T_n=\sum\limits_{i=1}^ni,\quad R_n=\sum\limits_{i=1}^ni^3. $$ Note that $T_n=\frac12n(n+1)$ and $R_n\leqslant\sum\limits_{i=1}^nn^3=n^4$, hence $$ \frac12\frac{n+1}n-\frac1{n^2}\leqslant S_n\leqslant\frac12\frac{n+1}n, $$ which implies that $$ \lim\limits_{n\to\infty}S_n=\frac12. $$
