most recent 30 from http://mathoverflow.net 2013-05-23T03:57:02Z http://mathoverflow.net/feeds/question/54758 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54758/does-the-following-series-converge Fabio 2011-02-08T13:14:37Z 2011-02-14T12:18:18Z

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

http://mathoverflow.net/questions/54758/does-the-following-series-converge/54792#54792 Lamine 2011-02-08T17:13:12Z 2011-02-14T12:18:18Z

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

http://mathoverflow.net/questions/54758/does-the-following-series-converge/54892#54892 Greg Kuperberg 2011-02-09T15:14:04Z 2011-02-09T15:14:04Z

<p>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 $$|\alpha n - k - \beta| &lt; 3/n$$ 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.</p>