Is $\sum_{n=1}^\infty a_n\sin(nx)$ converges on $[\varepsilon, 2\pi-\varepsilon]$?

Question

Let $a_n$, a sequence monotonically decreasing to $0$. Consider

$$\sum_{n=1}^\infty a_n\sin(nx)$$

Is the series converges uniformly on $[\varepsilon, 2\pi-\varepsilon]$? ($\varepsilon > 0$)

Basically we could use Dirichlet's test. We want to show that $\sum_{n=1}^\infty \sin (nx)$ is bounded. Indeed:

$$\sum_{n=1}^\infty \sin (nx) = \frac{i}{2}\left( \sum_{n=1}^\infty (e^{ix})^n + \sum_{n=1}^\infty (e^{-ix})^n\right) \le \frac{i}{2} \left( \frac{1}{1-e^{ix}} +\frac{1}{1-e^{-ix}} \right) \le \frac{1}{1-e^{i(2\pi-\varepsilon)}} <\infty$$

BUT, clearly, $$g(\frac{\pi}{2}) = \sum_{n=1}^\infty \sin \frac{n\pi}{2} = \infty$$

Where is the mistake?

I know you know this, but complex numbers dont have an order. — qqo, Jul 04 '15 at 18:49
Thank you all. By the way, why is the last inequality even holds? — Elimination, Jul 04 '15 at 19:37
@Nameless, it would be better if I used absolute value then? — Elimination, Jul 04 '15 at 19:44
I am not sure what you mean by "then". Anyways, your last series is alternating and divergent. — qqo, Jul 04 '15 at 20:10
Is there a differecen between convergence on $[\varepsilon, 2\pi - \varepsilon]$ for $\varepsilon > 0$ and $(\varepsilon, 2\pi)$? — MT_, Jul 04 '15 at 20:14

score 3 · Answer 1 · edited Apr 13 '17 at 12:20

We show that if $a_n$ is monotonically decreasing, then the series

$$\sum_{n=1}^{\infty}a_n\sin (nx)$$

is uniformly convergent on $[\epsilon,2\pi-\epsilon]$, for any fixed $\epsilon>0$.

To test the uniform convergence we use this answer in which I showed that

$$\begin{align} \left|\sum_{n=1}^N \sin(nx)\right| \le \frac12\left(1+\left|\cos (\frac{x}{2})\right|\right)\left|\csc\left(\frac{x}{2}\right)\right| \end{align}$$

Now, for any fixed $\epsilon$, we have for $x\in [\epsilon,2\pi-\epsilon]$,

$$\begin{align} \left|\sum_{n=1}^N \sin(nx)\right| &\le \frac12\left(1+\left|\cos (\frac{x}{2})\right|\right)\left|\csc\left(\frac{x}{2}\right)\right|\\\\ &\le \csc(\epsilon/2) \tag 1 \end{align}$$

For the Dirichlet test of $\sum_{n=1}^{\infty}a_n\sin (nx)$, we only require the following two conditions:

Condition $(1)$

The sequence $a_n$ decreases monotonically to zero.

Condition $(2)$

The partial sums $\sum_{n=1}^N \sin(nx)$ be bounded by a constant.

Condition $(1)$ is presumed while equation $(1)$ confirms Condition $(2)$.

Thus, if $a_n$ is monotonically decreasing, then for $x\in [\epsilon, 2\pi-\epsilon]$, for any fixed $\epsilon >0$, we have that

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^{\infty}a_n\sin (nx)\,\,\text{converges uniformly}}$$

As a side note, we have that the series $\sum_{n=1}^N \sin x \sin(nx)$ converges for $x\in [0,2\pi]$

We have

$$\begin{align} \left|\sum_{n=1}^N \sin x \sin(nx)\right| &\le \frac12\left(1+\left|\cos (\frac{x}{2})\right|\right)\left|\sin x\csc\left(\frac{x}{2}\right)\right|\\\\ &=\left(1+\left|\cos (\frac{x}{2})\right|\right)\left|\cos\left(\frac{x}{2}\right)\right|\\\\ &\le2 \tag 2 \end{align}$$

For the Dirichlet test of $\sum_{n=1}^{\infty}a_n\sin (x)\sin (nx)$, we only require the following two conditions:

Condition $(1)$

The sequence $a_n$ decreases monotonically to zero.

Condition $(2)$

The partial sums $\sum_{n=1}^N \sin (x) \sin(nx)$ be bounded by a constant.

Condition $(1)$ is presumed while equation $(2)$ confirms Condition $(2)$.

and we are done!

I am gonna have to favorite this question because of that box... — qqo, Jul 04 '15 at 20:01

score 2 · Answer 2 · answered Jul 04 '15 at 18:31

2

You last "clearly" is problematic. Because $g(\pi/2)$ is bounded between 0 to 1.

answered Jul 04 '15 at 18:31

Vim

13,905

Is $\sum_{n=1}^\infty a_n\sin(nx)$ converges on $[\varepsilon, 2\pi-\varepsilon]$?

2 Answers2