Find all $z \in \Bbb{C}$ such that $\sum_{n=1}^{\infty}\sin(n)z^n$ converges

Question

Find all $z \in \Bbb{C}$ such that $\sum_{n=1}^{\infty}\sin(n)z^n$ converges

I started by trying to compute the radius of convergence of the series, however Im absolutely stuck in finding $\limsup\sqrt[n]{|\sin(n)|}$.

I know that $\sin(n) \neq 1$ for every $n \in \Bbb{N}$ but I don't know how close can it get to $1$. Any hint?

$-1\leq \sin(n)\leq 1$. What does this say about the sum, or the limit? — pretzelman, Oct 07 '15 at 19:19
@pretzelman: not a lot: $-1 \le 1/n! \le 1$, but the power series for $e^z$ does converge. — Rob Arthan, Oct 07 '15 at 19:28
Because $\pi$ is irrational, $\limsup |\sin(n)|$=1. Hence the series converges for |z|<1 only. — A.S., Oct 07 '15 at 19:32
@D.Thomine If $|z|=1$, $|\sin(n)z^n|$ doesn't converge to 0, so the series can't converge. — A.S., Oct 07 '15 at 19:37
I showed here that $\lim\sup \sqrt[n]{|\sin(n)|} =1 $: https://math.stackexchange.com/questions/1460048/how-to-investigate-the-sup-inf-limsup-and-liminf-of-the-sequence/1460115#1460115 — marty cohen, Oct 07 '15 at 23:53

Rob Arthan · Answer 1 · 2015-10-07T19:51:34.593

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As (at the time I was typing) no-one else had posted a complete answer, I'm reinstating mine, with corrections due to insightful comments from A.S.

Because $\pi$ is irrational, the additive subgroup of $\Bbb{R}$ generated by $\Bbb{Z}$ and $\pi/2$ is dense in $\Bbb{R}$. This means that for any $N \in \Bbb{N}$ and $\epsilon > 0$ there is $n > N$ such that $|\sin(n) - 1| < \epsilon$. So $\sin(n)$ and hence $|\sin(n)z^n|$ does not tend to $0$ as $n$ tends to $\infty$ if $|z| \ge 1$, so $\sum_{n=1}^{\infty}\sin(n)z^n$ cannot converge. If $|z|< 1$, then $\sum_{n=1}^{\infty}\sin(n)z^n$ converges by the ratio test or comparison with a geometric series.

edited Oct 07 '15 at 19:51

answered Oct 07 '15 at 19:27

Rob Arthan

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Converges only when |z|<1. – A.S. Oct 07 '15 at 19:28
@A.S.: I think it's right now. Thanks for your comment. – Rob Arthan Oct 07 '15 at 20:00

Robert Israel · Accepted Answer · 2015-10-07T19:38:05.623

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EDITED $-1 \le \sin(n) \le 1$, so the series converges for $|z| < 1$.
Irrationality of $\pi$ implies $\sin(n)$ takes values arbitrarily close to $\pm 1$, so it diverges for $|z| \ge 1$.

edited Oct 07 '15 at 19:38

answered Oct 07 '15 at 19:33

Robert Israel

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what does more subtle mean? – Joaquin Liniado Oct 07 '15 at 19:36
$\sin n$ does not goes to zero as $n \to \infty$, so we can't have convergence when $z=1$. – abnry Oct 07 '15 at 19:37
Oops, I must have been thinking of something else for the "subtle" case. – Robert Israel Oct 07 '15 at 19:38
We only need to know $\sin n > 1/2$ for infinitely many $n,$ which is a bit more elementary than the irrationality business. – zhw. Oct 07 '15 at 19:51
@zhw.: the "irrationality business" is fundamental to my intuitions about the real line. What is the more elementary way of seeing that there are infinitely many $n$ with $\sin(n) > 1/2$? – Rob Arthan Oct 07 '15 at 20:03
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@RobArthan The interval $(\pi/6, 5\pi/6)$ has length $> 1$. – Daniel Fischer Oct 07 '15 at 20:09
@DanielFischer: interesting: the "irrationality business" seems fundamental to me, but the fact that $|\pi/6 - 5\pi/6| > 1$ is fundamental in a different kind of way. – Rob Arthan Oct 07 '15 at 21:39

score 1 · Answer 3 · answered Oct 07 '15 at 19:35

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Hint: Write as

$$\Im \left[ \sum_{n=1}^\infty (e^{i}z)^n \right].$$

answered Oct 07 '15 at 19:35

abnry

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$\sum \sin (n) z^n$ usually isn't real for non-real $z$. – Daniel Fischer Oct 07 '15 at 19:37
Whoops! Okay, the above works for $z$ real. – abnry Oct 07 '15 at 19:42

score 1 · Answer 4 · edited Apr 13 '17 at 12:21

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Look at my question and resolution, this might help you with your exercise:

Is series $\displaystyle\sum^{\infty}_{n=1}\frac{\cos(nx)}{n^\alpha}$, for $\alpha>0$, convergent?

It's almost the same situation, apply it to your problem.

edited Apr 13 '17 at 12:21

Community

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answered Oct 07 '15 at 19:47

Relure

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zhw. · Answer 5 · 2015-10-07T23:41:19.273

There is a more general result here, and it has nothing to do with the irrationality of $\pi.$ Claim: $\sum \sin (nx)z^n$ diverges for all $z,|z|=1$ and for all $x\in \mathbb {R}\setminus \pi\mathbb {Z}.$ I'll treat only the case $0< x \le 1.$ (See if you can fill in the details for the other values of $x.$) The proof is simple: As we view $e^{inx}$ marching around the unit circle infinitely many times, we see it has to land in the arc $A=\{e^{it}: \pi/4 <t < 3\pi/4\}$ infinitely many times. Why? Because the steps are of arc-length less than $1,$ the arc-length of $A$ is $\pi/2 > 1,$ and therefore we land in $A$ at least once every orbit. (Just like your momma said: You can't step over a puddle longer than your step.) So $\sin (nx)\ge 1/\sqrt 2$ for infinitely many $n,$ giving the result. There's surely a result here for any periodic function.

Find all $z \in \Bbb{C}$ such that $\sum_{n=1}^{\infty}\sin(n)z^n$ converges

5 Answers5