Limit of a sum of squared sines Following up on this question Nice Limit $\lim_{n\to\infty}\sum_{k=1}^{n} \sin^2\left(\frac{\pi}{n+k}\right)$ .
Fix a real $a$ and consider the sum:
$$S(n)=\sum_{k=1}^n \sin^2\left(\frac{n^a\pi}{n+k}\right)$$
In the linked question it is shown that for $a=0$ the sum converges to zero for large $n$. It is also easy to show that for $a=1$ the sum diverges, recognizing a Riemann sum. The question comes naturally: what happens for different $a$ ?
My thoughts for the moment: 


*

*For $0<a<1/2$ the sum converges to zero using the same arguments of the linked answer.

*For $1/2<a<1$ we can proceed like that. Using the Taylor series of $sin(x^2)$ we have that $sin(x^2) \ge x^2 - x^4/3$ so that:
$S(n) \ge \sum_{k=1}^n \frac{n^{2a}\pi^2}{(n+k)^2} \left( 1-\left( \frac{n^a \pi}{n+k}\right)^2 \frac{\pi^2}{3}
 \right)$ 
now $0 \le \left( \frac{n^a \pi}{n+k}\right) \le \left( \frac{n^a \pi}{2n}\right) \rightarrow_{n} 0$ if $a<1$
so we can safely maximize the term in the parenthesis with a constant $K_1>0$ and remove one term:
$S(n) > K_1 \sum_{k=1}^n \frac{n^{2a}\pi^2}{(n+k)^2} > K_2 \sum_{k=1}^n \frac{n^{2a}}{n^2} = K_2 n^{2a-1}$
, where $K_2$ is an other positive constant. The last term diverges for $a>1/2$.
Is it correct till now ?
I think then that for $a=1/2$ the limit converges to a non-trivial value.
I also conjecture that it diverges for $a>1$. 
For $a=1/2$ I have only naive arguments and for $a>1$ have no clue since the argument of the sine diverges and the sine oscillates for large arguments.
UPDATE: for $a>1$ I think one could try to define the angle $\alpha_k=\frac{n^a\pi}{n+k},k=1..n$. For large $n$ this rotates from $\frac{n^{a-1}\pi}{2}$ to $n^{a-1}\pi$. If one could prove that this angle remains "far" from $0$ and $\pi$ a sufficient number of times, divergence could than be proved. 
 A: For convenience, notations in the following proof are defined here.

1.$f(n)=\Theta(g(n))$ : there exist constans $A,B$ which are independent of $n$, such that
  $$\exists N,\forall n\geqslant N, Ag(n)\leqslant f(n)\leqslant Bg(n).$$
  2.$\psi'(x)$ : the second derivative of $\ln \Gamma(x)$.

There are some properties of $\psi'(x)$ in the link: http://dlmf.nist.gov/5.15.E8
For fixed $a\in (0,1)$, we have
$$\sin^2\frac{n^a\pi}{n+k}=\frac{n^{2a}\pi^2}{\left(n+k\right)^2}+\Theta\left(n^{4a-4}\right).$$
Thus as $n\to \infty$,
\begin{align*}
S(n)& =\sum_{k=1}^n\left(\frac{n^{2a}\pi^2}{\left(n+k\right)^2}+\Theta\left(n^{4a-4}\right)\right)\\
& =n^{2a}\sum_{k=n+1}^{2n}\frac{\pi^2}{k^2}+\Theta\left(n^{4a-3}\right)\\
& =n^{2a}\pi^2\left\{\psi'(n+1)-\psi'(2n+1)\right\}+\Theta\left(n^{4a-3}\right)\\
& =n^{2a}\pi^2\left(\frac{1}{2n}+\Theta\left(n^{-2}\right)\right)+\Theta\left(n^{4a-3}\right)\\
& =\frac{\pi^2}{2}n^{2a-1}+\Theta\left(n^{\max\{2a-2,4a-3\}}\right).
\end{align*}
Thence
$$\lim_{n\to \infty}S(n)=
\begin{cases}
0& \mathrm{if}\ a<\frac{1}{2},\\
\frac{\pi^2}{2}& \mathrm{if}\ a=\frac{1}{2},\\
+\infty& \mathrm{if}\ a>\frac{1}{2}.
\end{cases}
$$
