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As above, does $\sum_{n=1}^{\infty}(\sin{n})^n$ converge? And if so, to what value.

From calculating partial sums, it appears it might, but I'm not quite sure how to proceed from there.

David Reak
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    The matter is that $n mod \pi$ cannot be $0$ so $|\sin n|<1$. However it can go erratically very close to it. So the answer relates to the density near $1$ of $(\sin n)^n$ (but which I am not qualified to give) – G Cab Mar 10 '18 at 10:43
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    The answer is not obvious and is related to the irrationality measure of pi, see https://mathoverflow.net/questions/178811/lower-bound-on-the-irrationality-measure-of-pi for more explanations. – Did Mar 10 '18 at 10:44
  • @Did : in fact ! – G Cab Mar 10 '18 at 10:45
  • Assuming It would converge we could go for any $r>0$ grab an $N$ st for all $n\geq N$ we would have $|\sin n|^n < r$. On the other hand, if $\pi$ is normal, with probability one for any $s>0$ we could grap a $M>0$ such that $M>N$ and $|\sin M| > 1-s$. This leads to $N > \frac{\log r}{\log(1-s)}$ Question is, what to do with this result? – SK19 Mar 10 '18 at 11:32

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In Khinchin's little book on continued fractions the following theorem (Theorem 24) due to Chebyshev is given: For an irrational $\alpha$ and any $\beta$ there are infinitely many pairs of integers $x>0,y$ such that $|\alpha x-y-\beta|<\frac{3}{x}$.

We apply this for $\alpha=2\pi,\beta=-\frac{\pi}{2}$, and for any solution $x,y$ we get $$|1-\sin y|=|\sin(\alpha x-\beta)-\sin y|\leq|\alpha x-\beta-y|<\frac{3}{y}$$ (we use the inequality $|\sin a-\sin b|\leq|a-b|$, which follows from $|\sin'x|\leq 1$ and the mean value theorem), so $\sin y\geq 1-\frac{3}{y}$, $(\sin y)^y\geq\left(1-\frac{3}{y}\right)^y\to e^{-3}>0$ as $y\to\infty$. Therefore $(\sin n)^n$ doesn't converge to zero and the series doesn't converge.

Wojowu
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  • What if $\sin y < \frac{3}{y} +1$? – SK19 Mar 10 '18 at 11:43
  • From $|1-\sin y|<3/y$ both $\sin y>1-3/y$ and $\sin y<1+3/y$ follow. The latter is actually trivial, since $\sin y\leq 1$ always. – Wojowu Mar 10 '18 at 11:51
  • Okay, I got it, thanks. I'm still having problems wrapping my head around this proof, I mean I see why it is correct but I'm having trouble believing it, a rare occurrence for me – SK19 Mar 10 '18 at 11:54
  • So, am I right to generalize it like this? Show that $\sin(tn)^n$ doesn't converges for rational $t$. We would get $\sin(ty)\geq1-\frac{3}{ty}$ (since $ty$ is integer if $y$ is great enough) and conclude $\sin(ty)\geq \left(1-\frac{3}{ty}\right)^y\rightarrow e^{-3t^{-1}}$? – SK19 Mar 10 '18 at 12:01