# $S(x)=\sum_{n=1}^{\infty}a_n \sin(nx)$, $a_n$ is monotonic decreasing $a_n\to 0$: Show uniformly converges within $[\epsilon, 2\pi - \epsilon]$

$S(x)=\sum_{n=1}^{\infty}a_n \sin(nx)$, $a_n$ is monotonic decreasing $a_n\to 0$, when ${n \to \infty}$. I need to prove that for every $\epsilon >0$, the series is uniformly converges within $[\epsilon, 2\pi - \epsilon]$. Can I use Dirichlet and say that $\sum_{0}^{M} \sin (nx)< M$ for every x in the interval and since $a_n$ is uniformly converges to 0 ( uniformity since it does not depend on $x$), so the series is uniform convergent in this range?

In addition I need to prove that if $\sum_{n=1}^{\infty} a_n^2 = \infty$ so the series is not uniform convergent in $[0, 2 \pi]$, Since I know that from $n_0$ and on $a_n^2< a_n$ I used an inequality, again I'm not sure of that.

In the other hand, maybe I need to use Fourier series somehow.

Thanks for the help!

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Indeed, the fact that the sequence $\{\sum_{n=1}^N\sin(nx)\}$ is bounded will help us. More precisely, fix $\varepsilon>0$.

1. Show that we can find a constant $C=C(\varepsilon)$ such that for all $x\in [\varepsilon,2\pi-\varepsilon]$ we have $\left|\sum_{n=1}^N\sin(nx)\right|\leq C$ (to do that compute the sum).

2. Now we use Abel's tranform. We denote $s_N(x):=\sum_{n=0}^N\sin(nx)$ and $S_N(x):=\sum_{n=1}^Na_n\sin(nx)$. Then show that $$S_N(x)=\sum_{n=1}^Na_n(s_n(x)-s_{n-1}(x))=a_Ns_N(x)-a_1s_0(x)+\sum_{n=1}^{N-1}(a_n-a_{n-1})s_n(x).$$

3. Prove that the first two terms gives a uniformly convergent sequence (to $0$) whereas the third is a normally convergent sequence on $[\varepsilon,2\pi-\varepsilon]$, hence an uniformly convergent sequence on this interval.

For the second question, I will give you some steps:

1. Compute for fixed integers $m$ and $n$ $\int_0^{2\pi}S_n(x)S_m(x)dx$.
2. If we assume that $\{S_N\}$ is uniformly convergent on $[0,2\pi]$, then $$\lim_{n,m\to \infty}\int_0^{2\pi}S_n(x)S_m(x)dx=\int_0^{2\pi}S(x)^2dx.$$
3. Conclude.
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Thank you so much Davide! – Jozef Feb 12 '12 at 14:04
You're welcome. – Davide Giraudo Feb 12 '12 at 14:05
Did you manage to solve the problem? – Davide Giraudo Feb 12 '12 at 19:56
Yeah, I believe I did. :) – Jozef Feb 13 '12 at 5:59

You can use Dirichlet's Test$^\dagger$ as long as you can show that $$D_n(x)=\sum\limits_{k=1}^n \sin(kx)$$ is indeed uniformly bounded on $[\epsilon, 2\pi-\epsilon]$.

Dirichet himself did this as follows:

Using the formula $$2\sin(v)\sin(u)=\cos(v-u)-\cos (v+u),$$ for any $n$: \eqalign{ 2\sin(x/2) D_n(x)&= \sum_{k=1}^n\bigl[\, 2\sin(x/2)\sin(kx)\,\bigr]\cr &=\sum_{k=1}^n\Bigl[\, \cos \bigl(\,( k-{\textstyle{1\over2}})x\,\bigr) - \cos \bigl(\,(k+{\textstyle{1\over 2}})x\,\bigr)\,\Bigr]\cr &=\cos (x/2)-\cos \bigl(\,(n+\textstyle{1\over2})x\bigr). } So: $$\tag{1} |D_n(x)|= \biggl| {{\cos (x/2)-\cos \bigl(\,(n+{1\over2})x\,\bigr)}\over 2\sin(x/2) } \biggr| \le {{1\over |\sin(x/2)|} }.$$

Now, if $x\in [\epsilon, 2\pi-\epsilon]$, then $x/2\in[\epsilon/2, \pi-\epsilon /2]$ and it follows from inequality $(1)$ that $$|D_n(x)|\le {1\over \sin(\epsilon/2)}.$$

Dirichlet's Test:

Let $E\subset\Bbb R$ be a non-empty set and let $f_k$, $g_k$ be functions from $E$ to $\Bbb R$.

If $$\biggl|\sum\limits_{k=1}^n f_k(x)\,\biggr|\le M<\infty$$ for all positive integers $n$ and all $x\in E$, and if $g_k\searrow 0$ uniformly on $E$, then $\sum\limits_{k=1}^\infty f_k g_k$ converges uniformly on $E$.

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Interesting to see how Dirichlet did it. Another way: $\sin(kx)=Im(e^{ikx})$ so $\Sigma_{k=1}^N sin(kx)= Im \left(\sum_{k=1}^N e^{ikx} \right)$ which by Geometric Series formula is $Im ( \frac{1-e^{ix(N+1)}}{1-e^{ix}} )$ which, maximizing the top, is $\le \frac{2}{|1-e^{ix}|}$. Now $|1-e^{ix}|$ is continuous with zeros only at the endpoints of $[0, 2\pi]$. So on compact subsets, it attains a uniform minimum $>0$, which gives us a uniform finite bound $M_\epsilon$ on $[\epsilon, 2\pi-\epsilon]$. – Marcel T. Jul 22 '14 at 22:02