MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.