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Does the following series converge? $\sum_{n=1}^{\infty} \vert \sin n \vert ^{n}$

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Probably not. If instead of taking sin n, you take at the n'th stage sin x_n where x_n is uniformly distributed in [0, 2\pi] (the density of n 'mod' 2*\pi should behave the same), you get that |x_n - \pi/2| < 1/n infinitely often. Each such occurrence contributes O(1) to the sum. Also, not sure MO is the right venue for these type of questions. –  Or Zuk Feb 8 '11 at 13:30
It's even worse than that. $\sin (\pi/2 \pm \epsilon) \geq 1 - \epsilon^2/2$. So, just the weaker bound $|n - (2k+1)\pi/2| < 1/\sqrt{n}$ gives $(\sin n)^n > e^{-1/2}$. I would guess that it is not too bad to show that $n$ is infinitely often this close to an odd multiple of $\pi/2$, but I don't see the details right now. –  David Speyer Feb 8 '11 at 13:55
"Chebyshev's Theorem" (Khinchin's continued fraction book) says that for arbitrary irrational $\alpha$ and real $\beta$ the inequality $|\alpha x-y-\beta|<3/x$ has infinitely many integer solutions. From this follows easily that $n$ is infinitely often as close to an odd multiple of $\pi/2$ as David's argument requires. –  SJR Feb 8 '11 at 14:41
this looks too much like homework to me. –  Igor Rivin Feb 8 '11 at 15:20
It reminds me an other innocent looking series: $$\sum\frac{(-1)^n}{\ln n+\cos n}.$$ It converges, but to prove it you need to know a bit about rational approximations of $\pi$. –  Anton Petrunin Feb 8 '11 at 17:16
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2 Answers

up vote 5 down vote accepted

The question has basically been answered in the comments by David Speyer and SJR. It is a theorem of Chebyshev that that for any irrational $\alpha$ and any real $\beta$, the inequality $$|\alpha n - k - \beta| < 3/n$$ has infinitely many solutions. In particular, take $\alpha = 1/(2\pi)$ and $\beta = \frac12$. Then one gets that $n$ is so close to an odd multiple of $\pi$ that $|\sin n|^n$ converges to 1 for these values. Even if you took $|\sin n|^{n^2}$, these values would be bounded away from 0. Certainly if the terms of a series do not converge to 0, then the series does not converge.

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Since $\pi$ is transcendental (so also $\frac{\pi}{2}$ and $\frac{3\pi}{2}$), $\forall n \in \mathbb{N} , |\sin{n}|<1$. In another hand, $\sum_{n=2}^\infty|\sin{n}|^n <\sum_{n=2}^\infty|\sin{n}|^2$ which converges (because $\sum_{n=1}^\infty a^n$ converges if $|a| < 1$.

So, $\sum_{n=2}^\infty|\sin{n}|^n$ converges.

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This doesn't work because the comparison to a geometric series is not apt; since $|\sin(n)|$ could be arbitrarily close to 1 infinitely often (so that $|\sin(n)|^n$ could also be arbitrarily close). –  Stanley Yao Xiao Feb 8 '11 at 18:02
The reply is not correct because $\vert \sin n \vert^{2}$ has a fixed exponent and a basis dependent on $n$, the latter being infinitely many time close to 1, therefore the comparison with the geometric series has no meaning. Surely $\sum \vert \sin n \vert^{2}$ diverges, because $\vert \sin n \vert$ is infinitely many times close to 1, therefore $\vert \sin n \vert^{2}$ is infinitely many times bigger than $\frac{1}{2}$, so that the series diverges. –  Fabio Feb 8 '11 at 18:10
I'm ashamed of this answer. That's what happens when I don't sleep enough. –  Lamine Feb 14 '11 at 12:16
It may be wrong, but -9 votes? Really? –  Yemon Choi Feb 14 '11 at 12:25
Scratching out the wrong answer is a classy move. –  userN Feb 14 '11 at 16:45
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