Does $\sum_n |\sin n|^{cn^2}$ converge?

So I recently asked a question about convergence of $\sum_n |\sin n|^{cn}$ for arbitrary $c > 0$ and it turns out that the terms of the series don't even converge, for any $c > 0$, so the series is always divergent. But what about $\sum_n |\sin n|^{cn^2}$ for $c > 0$? Are there $c$ so that the series converges, and if there are $c > 0$ such that the series diverges, do the terms of the series still converge?

If that's too hard, what if we replace $n^2$ in the exponent by $n^\alpha$ for some different $\alpha > 1$? Which $\alpha$ do we know the answer for?

Solution 1:

It is possible to prove that $|\sin n|^{n^2}$ does not have a limit. We do this by showing that there are subsequences converging to different limits.

First, note that there is a subsequence converging to zero. This is because there are infinitely many $n$ with $|\sin n|\leq 1/2$.

For the other subsequence, consider for any integer $k$, $$ \begin{align} |\sin n| &= |\sin(n -k\pi)| \end{align} $$ If we find a positive constant $C_1$ and infinitely many $n$, $k$ satisfying $$ \left| n - k\pi - \frac{\pi}2 \right|\leq \frac {C_1} n, \ \ \ (*) $$ Then we obtain a positive constant $C_2$ and infinitely many $n$ with $$ |\sin n|\geq 1-\frac {C_2}{n^2}. $$ Then by taking $n^2$-th power, we have $$ |\sin n|^{n^2} \geq \left(1-\frac{C_2}{n^2}\right)^{n^2} \rightarrow e^{-C_2} \ \mathrm{as } \ n\rightarrow\infty. $$

Now, we achieve ( *) by considering the continued fraction for $\pi/2$. Note that ( *) is equivalent to $$ \left|\frac n{2k+1} - \frac{\pi}2 \right| \leq \frac{C_3}{n^2} $$ for some positive constant $C_3$.

To achieve ( *), let $n$-th convergent $p_n/q_n$ of the continued fraction of $\pi/2$ to satisfy $q_n$ is odd infinitely often. This is possible since $p_nq_{n-1}-p_{n-1}q_n=(-1)^n$. By this identity, no two consecutive denominators $q_n$'s are even.

Hence, your series $\sum |\sin n|^{cn^2}$ is divergent, since the terms diverge.