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I have been trying to find the convergence (and value) of this infinite sum: $$\sum_{n=1}^\infty \frac{\sin(n)^n}{\cos(en)}$$ The partial sums behave relatively unpredictably and at some point become very hard to evaluate because of the exponent on the $\sin$, but from what I've seen I can't tell if it converges or not, although since the top is raised to the power $n$ it feels like it should. I tried the ratio test but that really doesn't lead anywhere, and applying L'Hôpital's rule doesn't help evaluate the limit. WolframAlpha can't give any useful info about the sum or its convergence either.

TigerGold
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  • Surely you understand that, for all x, sin(x) lies between -1 and 1. So as n goes to infinity, [tex]sin^n(n)[/tex] goes to 0. – user247327 Oct 25 '20 at 23:17
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    @user247327 yes of course, but the bottom is occasionally very close to zero, which means that the partial sums jump by (sometimes) enormous amounts at certain points. – TigerGold Oct 25 '20 at 23:19
  • Is the $e$ in the argument of $\cos$ correct or it is just a typo? If it is not a typo, is it the Euler's number or is it an independent variable? – the_candyman Oct 25 '20 at 23:34
  • @the_candyman it's not a typo and it's Euler's number. – TigerGold Oct 25 '20 at 23:39
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    @user247327 - Your reasoning is wrong. $1-\tfrac1n$ also lies between $1$ and $-1$, but $(1-\tfrac1n)^n$ does not converge to $0$. (Indeed it converges to $e^{-1}$.) – mr_e_man Oct 25 '20 at 23:40
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    By Dirichlet's theorem on Diophantine approximation, there are infinitely many integers $q$ such that $q\pi$ is within $\frac1q$ of an integer $n$. For such $n$, $|\sin n|^n$ is at least $|1-\frac1{2q^2}|^n \ge |1-\frac1{2q^2}|^{q\pi +1}$, which tends to $1$ as $q$ tends to infinity. So unless $en$ is extremely non-uniformly distributed modulo $\pi$ on this special subsequence, the sum diverges because its summand does not tend to $0$. Perhaps there's some quantitative measure of the linear independence of $1$, $\frac1\pi$ and $\frac1{e\pi}$ that could formalize this argument. – Greg Martin Oct 25 '20 at 23:47
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    @GregMartin - Doesn't that already answer the question? $\frac{1}{|\cos(en)|}\geq1$, so for that special subsequence, $$\frac{|\sin n|^n}{|\cos(en)|}\geq|\sin n|^n\to1$$ which doesn't approach $0$. – mr_e_man Oct 25 '20 at 23:58
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    In fact your series may be undefined! It's not known whether $\frac e\pi$ is rational, so it's possible that $\cos(en)=0$ for an integer $n$. – mr_e_man Oct 26 '20 at 00:16
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    @mr_e_man for the sake of the question we should assume $e/\pi$ is irrational, because otherwise the question is kind of pointless and also it seems (intuitively to me lol) that it almost certainly is. – TigerGold Oct 26 '20 at 00:21
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    @GregMartin actually you want $n$ to be close to $\pi q/2$ (with $q$ odd) rather than $\pi q$ to have $|\sin(n)|$ close to $1$ (not $0$). – Robert Israel Oct 26 '20 at 01:08

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