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I saw this series incidentally: $$\sum_{n=1}^\infty (\sin n)^n $$ Result from WolframAlpha seems to say the series diverges but I don't know how to prove it. Thanks for any help!

Hanul Jeon
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  • If we show $\left{{\frac{n}{2\pi}}|n>N, n\in \mathbb{N}\right}$ is a dense set in $(0;1)$ $\forall N\in \mathbb{N}$ then $\left(\frac{\sin^{-1}(\sqrt[n]{\frac{1}{2}})}{2\pi};\frac{\pi-\sin^{-1}(\sqrt[n]{\frac{1}{2}})}{2\pi}\right)$ always contains a number $x$ for which $\sin^n (2\pi x)>\frac{1}{2}$ so $\lim\limits_{n\to\infty} \sin^n(n)\neq 0$. – Alexey Burdin Apr 27 '15 at 02:56

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In order to have a series converge, you must have the summands approach $0$ as a limit. Although intuitively you might think that $-1<\sin(n)<1$ for all $n\in\mathbb{N}$ and therefore $(\sin(n))^n$ acts like $r^n$ with $-1<r<1$ and will converge, this seems to fail.

Lemma: For each $x\in[-1,1]$, there exists a subsequence of $A_n=\sin(n)$ such that $\lim\limits_{n_j\to\infty} A_{n_j}=x$

For proof of the lemma, see sin(n) subsequence limits set

So then, there exist infinitely many $n$ for which $\sin(n)\geq 1-\frac{1}{n}$, and therefore infinitely many $n$ for which $(\sin(n))^n>\frac{1}{e}$. Therefore the sequence of partial sums will always have some sudden jumps and cannot converge.

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JMoravitz
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  • JMoravitz> "So then, there exist infinitely many $n$ for which $\sin(n)≥1−\frac1n$." Does this follow from the lemma? I doubt. Then $\forall n$ there exist infinitely many $n_k$ for which $\sin(n_k)≥1−\frac1n$. But what follows from this? – grizzly Dec 01 '17 at 15:05
  • @grizzly It has been quite a while since I visited this question. Lets see... My claim was that the lemma implied that there are infinitely many $n$ for which $\sin(n)\geq 1-\frac{1}{n}$ (not that given a specific $n$ there are infinitely many $n_k$ such that $\sin(n_k)\geq 1-\frac{1}{n}$). If this claim happened to be true, then we could show that there is an infinite subsequence of $a_n=\sin(n)^n$ for which $a_n>\frac{1}{e}$ as we would have had for those values of $n$ that $\sin(n)^n\geq (1-\frac{1}{n})^n>\frac{1}{e}$. This would imply that $a_n$ does not converge to zero. – JMoravitz Dec 01 '17 at 19:13
  • Knowing that $\sin(n)^n$ does not converge to zero, this implies by the $n$'th term divergence test that the series could not converge either. (If a series converges, then the summands must converge to zero. If the summands do not converge to zero, then the series diverges). Looking back on this, I feel like I thought my claim was obvious but it is not obvious to me now. I still feel like it should intuitively be true, but I would have to think on it further to prove it. Thankfully, there is already another linked answer which shows $\sin(n)^n$ diverges, implying the final step. – JMoravitz Dec 01 '17 at 19:16
  • I will delete my answer after a few hours once you've had a chance to read this since I cannot back up that claim. – JMoravitz Dec 01 '17 at 19:20
  • Thank you for your attention. I understood your comment. I'm glad if my comment was useful. – grizzly Dec 01 '17 at 19:32