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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.

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Maybe these answers could help you:math.stackexchange.com/questions/109029/… , math.stackexchange.com/questions/2270/…. –  Davide Giraudo Apr 2 '12 at 21:14
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3 Answers

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.

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Very nice proof! –  Salech Alhasov Apr 2 '12 at 22:53
    
@SalechAlhasov: actually, reading Robert Israel's answer, I notice that I can make things better. I will update my answer. –  robjohn Apr 2 '12 at 23:10
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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 $|\sin n|^n > 1 - \epsilon$.

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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 2 '12 at 22:51
    
And reading your answer, I see that I can raise that to $1$ :-) –  robjohn Apr 2 '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 $1$.

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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 $0$. –  Henning Makholm Apr 2 '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 2 '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. –  Henning Makholm Apr 2 '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 2 '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 2 '12 at 23:06
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