I'm almost sure that the series $\sum \sin^n(n)$ is not convergent, but lack proof. Thank for any help.
Is the series $\sum \sin^n(n)$ divergent?
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5Maybe 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. – 2012-04-02
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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0@SalechAlhasov: actually, reading Robert Israel's answer, I notice that I can make things better. I will update my answer. – 2012-04-02
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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0And reading your answer, I see that I can raise that to $1$ :-) – 2012-04-02
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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0Well, I did start my answer with the word "Hint." @robjohn – 2012-04-02