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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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    @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.