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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?

Thanks

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    Note that $R \geq 1$ since $$\left| \sum_{n=0}^{\infty} x^n \sin(nx) \right| \leq \sum_{n=0}^{\infty} |x|^n |\sin (nx)| \leq \sum_{n=0}^{\infty} |x|^n.$$2012-02-07
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    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.2012-02-07
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    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.2012-02-07
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    @JavaMan: I'm not sure I understand your conclusion from these inequalities, can you extend it? Thnaks!2012-02-07
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    @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.2012-02-07
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    @JavaMan: Ok, but then you wrote that $R \geq 1$, while the second series has a radius smaller than one. Is this a mistake?2012-02-07
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    @Jozef: It wasn't a mistake, I just wasn't being specific enough I guess. I meant the radius of convergence of the series $\sum x^n \sin(nx)$ is at least as big as $1$.2012-02-07

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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.