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Given this functions series : $\sum_{n=0}^{\infty}x^n\sin (nx)$, I need to find the ratio where it converges. I don't see how can I change it into a form where I'll be able to use Cauchy-Hadmard or d'Alambert theorems into order to find R, the radius of convergence.

Any suggestions?


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Note that $R \geq 1$ since $$\left| \sum_{n=0}^{\infty} x^n \sin(nx) \right| \leq \sum_{n=0}^{\infty} |x|^n |\sin (nx)| \leq \sum_{n=0}^{\infty} |x|^n.$$ – JavaMan Feb 7 '12 at 19:38
Rewrite it as the imaginary part of $\sum_{n=0}^\infty (xe^{ix})^n$, perhaps? Then it clearly converges for (real) $|x|<1$. But does divergence of the sum mean divergence of the imaginary part of the sum? Not sure. – Thomas Andrews Feb 7 '12 at 20:02
Lemma: If $\frac{x}\pi$ is irrational, then there are infinitely many positive integers $n$ such that $|\sin nx| > \frac{1}{2}$. If you can prove that, then for any $R>1$ you can find such an x in $(1,R)$, and the sequence $x\sin nx$ cannot converge to zero, and hence its sum cannot converge. – Thomas Andrews Feb 7 '12 at 20:36
@JavaMan: I'm not sure I understand your conclusion from these inequalities, can you extend it? Thnaks! – Jozef Feb 7 '12 at 20:59
@Jozef: All I am using is that if $\sum a_n \leq \sum b_n$, and $\sum b_n$ converges, then so does $\sum a_n$. This shows that the radius of convergence of the first sum is at least as large as the radius of convergence for the second sum. – JavaMan Feb 7 '12 at 21:08
up vote 5 down vote accepted

By comparison with $\sum |x|^n$, our series converges absolutely if $|x|<1$. Let's see why things go bad for most $|x|\ge 1$. Except when $x$ is of the form $k\pi$, the terms do not have limit $0$.

To do this, you will have to show that except in the case when $x$ is an integer multiple of $\pi$, we can find a positive $\alpha$ such that infinitely many integers $n$, $|\sin(nx)|>\alpha$. Hint: Suppose that by bad luck $\sin(nx)$ is awfully close to $0$. Show that $\sin((n+1)x)$ isn't.

Remark: I am not sure about the use of the term convergence radius. With power series, we have divergence if $|x|$ is greater than the convergence radius. Here we mostly have divergence, but the points $k\pi$ are exceptional.

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