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I'm trying to show the series ${\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\sin\left(nx\right)}$ converges for all $x\in\left[0,2\pi\right]$. Using Dirchlet's Test it suffices to show that the series of partial sums of ${\displaystyle \sum_{n=1}^{\infty}\sin\left(nx\right)}$ is bounded but I can't seem to manage to show that. I saw a solution that uses some complex number identities but I really would prefer to avoid using complex numbers.

Help would be appreciated!

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Compare $\dfrac{1}{\sqrt{n}}\sin(nx)$ with $\dfrac{1}{n}$ and use the comparation test; also note that $|\sin(nx)|\leq1$. – Marra Feb 14 '13 at 17:08
See here. Note this will show convergence on $(0,2\pi)$; of course, the series converges at $0$ and $2\pi$ as well. – David Mitra Feb 14 '13 at 17:10
$\frac{1}{n}$ does not converge though... – Serpahimz Feb 14 '13 at 17:10
Yes, my bad. Sorry! – Marra Feb 14 '13 at 17:16
up vote 4 down vote accepted

Oh that is a nice one. Taking $$ \sin(kx)=\frac{\cos\bigl((k-\frac{1}{2})x\bigr)-\cos\bigl((k+\frac{1}{2})x\bigr)}{2\sin\frac{x}{2}}$$ you can see that $$\left|\sum_{k=1}^n \sin(kx)\right|\leq\left|\frac{1}{\sin\frac{x}{2}}\right|$$ for all $n \in \mathbb{N}$ if $x\neq 2 \pi k $ with $k \in \mathbb{Z}$

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(+1): So, for this problem, that's all you need in order to show convergence for $x\in(0,2\pi)$, and it's immediate for $x=0,x=2\pi$. – Cameron Buie Feb 14 '13 at 17:26
Using Trig identities has always been my weak spot. Very nice solution though, thanks to all the helpers! – Serpahimz Feb 14 '13 at 17:48
Im not following. You used a telescoping series? Because the limit of the second term does not converge. – CogitoErgoCogitoSum Feb 14 '13 at 17:55
To show the inequality written above he indeed used the fact it's a telescoping sum. Note that the series $sin(kx)$ does not converge but for Dirchlet's Test you only need to show that its partial sums are bounded, which is what was shown here. – Serpahimz Feb 14 '13 at 17:59

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