<s>[INST] Show that $\lim\limits_{n \to +\infty}(\sin(\frac{1}{n^2})+\sin(\frac{2}{n^2})+\cdots+\sin(\frac{n}{n^2})) = \frac{1}{2}$ 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.
 [/INST] Because $\sin x$ is convex on $x \in (0, \pi)$ then the sum of $\sin (kx)$ inside that interval will be not less than $n$ times the linear average on the secant $(\sin x+sin (nx))/2$ and not greater than $n$ times the linear average on the tangent $n \sin((n+1)/2 x)$, so in this case
$$
\eqalign{
  & n\left( {{{\sin \left( {1/n^{\,2} } \right) + \sin \left( {n/n^{\,2} } \right)} \over 2}} \right) \le x_{\,n}  =   \cr 
  &  = \sin \left( {{1 \over {n^{\,2} }}} \right) + \sin \left( {{2 \over {n^{\,2} }}} \right) +  \cdots  + \sin \left( {{n \over {n^{\,2} }}} \right)
 \le n\sin \left( {{{\left( {n + 1} \right)/2} \over {n^{\,2} }}} \right) \cr} 
$$
</s>