Solving $\lim_{n \rightarrow \infty} \sum_{r=1}^{n} \sin {\frac{r}{n^2}}$ in a "non-traditional" way $$\lim_{n \rightarrow \infty} \sum_{r=1}^{n} \sin {\frac{r}{n^2}}$$ I know how to solve this by using squeeze theorem.
I solved it by converting  $\sin {\frac{r}{n^2}} \rightarrow \frac{1}{n} \cdot n \sin{ \left( \frac{r}{n} \cdot \frac{1}{n} \right) }$
Thus the problem became to solve $$\lim_{n \rightarrow \infty} {n} \cdot \int_0^1 \sin {\left( \frac{x}{n}\right)}dx$$
Which on expanding the integral and changing the variable, gives us $$\lim_{t \rightarrow 0^+} \frac{1-\cos t }{t^2}$$ Which gives the correct answer of $\frac{1}{2}$.
Now I want to know whether this thing that I did here is correct or not. (correct in the sense of me able to use this type of technique all the time without anything breaking!)
 A: Actually, we can compute the sum $\sum_{k=1}^{n}\sin\left(\frac{k}{n^{2}}\right)$ exactly. Let $S_{n}=\sum_{k=1}^{n}\cos\left(\frac{k}{n^{2}}\right)$
and $T_{n}=\sum_{k=1}^{n}\sin\left(\frac{k}{n^{2}}\right)$. Consider
\begin{eqnarray*}
S_{n}+iT_{n} & = & \sum_{k=1}^{n}\cos\left(\frac{k}{n^{2}}\right)+i\sin\left(\frac{k}{n^{2}}\right)\\
 & = & \sum_{k=1}^{n}\exp\left(\frac{i}{n^{2}}\cdot k\right)\\
 & = & \sum_{k=1}^{n}\left[\exp\left(\frac{i}{n^{2}}\right)\right]^{k}.
\end{eqnarray*}
Denote $z=\exp\left(\frac{i}{n^{2}}\right).$ By sum of G.P. formula,
we have that $S_{n}+iT_{n}=\frac{z(1-z^{n})}{1-z}.$ By writing out
$z$ and $z^{n}$ explicitly, we have $z=\cos(\frac{1}{n^{2}})+i\sin(\frac{1}{n^{2}})$
and $z^{n}=\cos(\frac{1}{n})+i\sin(\frac{1}{n})$. Therefore, the
expression $\frac{z(1-z^{n})}{1-z}$ can be simplified and expressed
in the form $X+iY$. (I do not have a piece of paper to carry out
the exact computation). Lastly, comparing the real and imaginary parts,
we obtain $T_{n}=Y$.
