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I'm almost sure that the series $\sum \sin^n(n)$ is not convergent, but lack proof. Thank for any help.

Luciano Battaia
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    Maybe these answers could help you:http://math.stackexchange.com/questions/109029/convergence-divergence-of-infinite-series-sum-n-1-infty-frac-sinn2 , http://math.stackexchange.com/questions/2270/convergence-of-sum-limits-n-1-infty-sinnk-n. – Davide Giraudo Apr 02 '12 at 21:14

3 Answers3

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Using a continued fraction approximation, we know that for any positive integer $N$, we can find integers $q>N$ and $p$ so that $$ \left|p-q\frac\pi2\right|<\frac1q\tag{1} $$ It is also true that no two consecutive denominators can share a common factor. Therefore, if one approximation has an even denominator, the next must be odd. So assume that $q$ is odd and $(1)$ is true. Then, since $\sin\left(q\frac\pi2\right)=(-1)^{(q-1)/2}$ and $\cos\left(q\frac\pi2\right)=0$ the Maclaurin Series yields

$$ \begin{align} (-1)^{(q-1)/2}\sin(p) &\ge1-\frac12\left(p-q\frac\pi2\right)^2\\ &\ge1-\frac{1}{2q^2}\tag{2} \end{align} $$ Thus, for continued fraction approximations $\frac{p}{q}$ to $\frac\pi2$ with odd denominators $|\sin(p)|\ge1-\frac{1}{2q^2}$. Taking the $\liminf$ as $q\to\infty$ yields $$ \begin{align} \liminf_{p\to\infty}|\sin(p)|^p &\ge\lim_{p\to\infty}\left(1-\frac{1}{2q^2}\right)^p\\ &=\lim_{q\to\infty}\left(1-\frac{1}{2q^2}\right)^{q\pi/2}\\ &=1\tag{3} \end{align} $$ Inequality $(3)$ implies that $$ \limsup_{n\to\infty}|\sin^n(n)|=1\tag{4} $$ Since the terms do not tend to $0$, $$ \sum_{n=0}^\infty\sin^n(n) $$ does not converge.

robjohn
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For any irrational $r$ (in particular $\pi/2$) there are infinitely many fractions $p/q$ such that $|r - p/q| < 1/q^2$. IIRC we can specify that $q$ is odd, perhaps at the cost of changing $1/q^2$ to $c/q^2$ for some constant $c$. Taking $r = \pi/2$, this says $|q \pi/2 - p| < c/q$, and then $|\sin p| > 1 - c^2/(2 q^2)$ and $|\sin p|^p > (1-c^2/(2 q^2))^p$. Now as $q \to \infty$ with $p \approx q \pi/2$, $(1 - c^2/q^2)^p \to 1$. Thus for any $\epsilon > 0$ there are infinitely many $n$ with $|\sin n|^n > 1 - \epsilon$.

Robert Israel
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    Since consecutive denominators are relatively prime, if one denominator is even, the next must be odd. Therefore, we can keep $1/q^2$, just skip an approximation once in a while. I use this in my answer to get $$\limsup_{n\to\infty}\left|\sin^n(n)\right|\ge e^{-\pi/2}$$ An interesting use for $i^i$. – robjohn Apr 02 '12 at 22:51
  • And reading your answer, I see that I can raise that to $1$ :-) – robjohn Apr 02 '12 at 23:11
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Hint: If $\frac{n}{k}$ is a "good" approximation to $\frac{\pi}2$ and $k$ is odd and large, then $|\sin^n n|$ is close to $1$.

Thomas Andrews
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    It's not quite obvious, though, that there can be enough $n/k$ that approximate of $\pi/2$ well enough to overcome the tendency of the $n$th power to draw an almost-$1$ down towards $0$. – hmakholm left over Monica Apr 02 '12 at 21:39
  • Not clear what you mean, @HenningMakholm, but there are infinitely many such $n/k$, so $\sin^n n$ cannot converge to zero, so the sum cannot converge. Showing that there are infinitely many $n/k$ is as easy as the continued fraction, and a simple parity proof that, for any irrational, there are infinitely many convergents of its continued fraction with odd denominator. (Basically, at least one of every two consecutive convergents has an odd denominator.) – Thomas Andrews Apr 02 '12 at 21:46
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    What I mean is just what I wrote -- that it is not obvious to me. The larger $n$ is, the better must the approximation $n\approx k\frac{\pi}{2}$ be in order to guarantee $|\sin^n n|>a$ for any fixed $a>0$. It is not clear to me how to estimate the various growth rates such as to establish the result. Perhaps I'm just tired. – hmakholm left over Monica Apr 02 '12 at 21:52
  • The continued fraction convergents, $p_n/q_n$ for a real number $\alpha$ have the property that $|p_n/q_n-\alpha|<1/q_n^2$. When $\alpha$ is irrational, there are infinitely many convergents, and infinitely many with $q_n$ odd. Then all you need to show is that this is "close enough." But if you haven't seen continued fractions, this is definitely not obvious. @HenningMakholm – Thomas Andrews Apr 02 '12 at 22:25
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    @Thomas: I think that Henning was hoping to see some of that in your proof. I know about continued fractions, and I still needed to write it out to make sure everything worked out right. In any case, it would be nice to at least reference "continued fraction approximations" so that those who might not know about continued fractions would have a starting point. – robjohn Apr 02 '12 at 23:06
  • Well, I did start my answer with the word "Hint." @robjohn – Thomas Andrews Apr 02 '12 at 23:19