TITLE: Show that $\lim\limits_{n \to +\infty}(\sin(\frac{1}{n^2})+\sin(\frac{2}{n^2})+\cdots+\sin(\frac{n}{n^2})) = \frac{1}{2}$
QUESTION [13 upvotes]: Show that the sequence defined as
$$x_n = \sin\left(\frac{1}{n^2}\right)+\sin\left(\frac{2}{n^2}\right)+\cdots+\sin\left(\frac{n}{n^2}\right)$$
converges to $\frac{1}{2}$.
My attempt was to evaluate this limit by using squeeze theorem. I managed to show that $x_n < \frac{n+1}{2n}$ by using $\sin(x) < x$, but I haven't been able to find a sequence smaller than $x_n$ that also converges to $\frac{1}{2}$. I tried showing by induction that $x_n > \frac{1}{2}-\frac{1}{n}$, but I got nowhere with that.
Any help would be appreciated.

REPLY [3 votes]: Another method, because it wasn't mentioned, is using Lagrange's trigonometric identities (which should not be difficult to prove using complex numbers, like here)
$$\sum _{k=1}^{n}\sin(k\theta )=
{\frac {1}{2}}\cot {\frac {\theta }{2}}-{\frac {\cos \left(\left(n+{\frac {1}{2}}\right)\theta \right)}{2\sin \left({\frac {\theta }{2}}\right)}}=...$$
with $\theta=\frac{1}{n^2}$. Given $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$
$$...={\frac {1}{2}}\cot {\frac {\theta }{2}}-\frac{1}{2}\left(\cos\left(n\theta\right)\cot\frac{\theta}{2}-\sin(n\theta)\right)=\\
\frac{1}{2}\cdot\left(\sin(n\theta)+\cot\frac{\theta}{2}\cdot\left(1-\cos\left(n\theta\right)\right)\right)=\\
\frac{1}{2}\cdot\left(\sin\frac{1}{n}+\cot\frac{1}{2n^2}\cdot\left(1-\cos\frac{1}{n}\right)\right)=...$$
considering that $1-\cos{a}=2\sin^2\frac{a}{2}$
$$...=\frac{1}{2}\cdot\left(\sin\frac{1}{n}+2\cdot\cot\frac{1}{2n^2}\cdot \sin^2\frac{1}{2n}\right)=...$$
also considering $\color{red}{\lim\limits_{x\to0}\frac{\sin x}{x}=1}$
$$...=\frac{1}{2}\cdot\left(\sin\frac{1}{n}+2\cdot\cos\frac{1}{2n^2}\cdot \frac{\frac{1}{2n^2}}{\sin\frac{1}{2n^2}}\cdot \frac{\sin^2\frac{1}{2n}}{\frac{1}{4n^2}}\cdot\frac{\frac{1}{4n^2}}{\frac{1}{2n^2}}\right)=\\
\frac{1}{2}\cdot\left(\sin\frac{1}{n}+\cos\frac{1}{2n^2}\cdot \color{red}{\left(\frac{\frac{1}{2n^2}}{\sin\frac{1}{2n^2}}\right)}\cdot \color{red}{\left(\frac{\sin\frac{1}{2n}}{\frac{1}{2n}}\right)^2}\right)\to \frac{1}{2}, n\to\infty$$