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I'm trying to understand why the series $$\sum_{n=1}^\infty \frac {\sin (nx)}{n^{\alpha}}$$ converges for $\alpha > 0$.

At the end of the prof for $0 < \alpha \le 1 $ it is not clear to me two passages.

My book says that this result is obvious for $\alpha > 1$ because the series converges absolutely.

$|\sin(nx)| \le 1 \Rightarrow |\frac {\sin (nx)}{n^{\alpha}}| \le \frac {1}{n^{\alpha}} \forall x \in R$.

For $\alpha >1 \sum_{n=1}^\infty \frac {1}{n^{\alpha}}$ converges.

So $\sum_{n=1}^\infty \frac {\sin (nx)}{n^{\alpha}} \le \sum_{n=1}^\infty \frac {1}{n^{\alpha}}$ and for the comparison test $\sum_{n=1}^\infty \frac {\sin (nx)}{n^{\alpha}}$ converges.

For $0 < \alpha \le 1 $ we can use the Dirichlet criterion. To do this we have to prove that $\sigma_n=\sum_{k=1}^n \sin (kx)$ is limited.

Once we have done this, the sequence $\left \{ \frac{1}{n^{\alpha}}\right \}$ is positive and decreasing and we can say that $\sum_{n=1}^\infty \frac {\sin (nx)}{n^{\alpha}}$ respects the Dirichlet criterion and it converges.

I have not understood the proof of the limitation of $\sigma_n=\sum_{k=1}^n \sin (kx)$.

In my book it suggests to analyze the complex sequence $e^{ikx}= \cos (kx)+i\sin (kx)$

with $r_n=\cos (kx)$ and $s_n=\sin (kx)$.

$\sum_{n=1}^\infty e^{ikx}= \frac{e^{i(n+1)x}-1}{e^{ix}-1}$

and then$|\sum_{n=1}^\infty e^{ikx}|=\sum_{n=1}^\infty \sqrt{r_n^2+s_n^2} =\sum_{n=1}^\infty \sqrt{{\cos(kx)}^2+{\sin(kx)}^2} \le \frac {2}{|e^{ix}-1|} $

Why this last passage and $|e^{i(n+1)x}-1| \le 2$? and then the book continues saying that for $x \ne 2k \pi$ both the sequences $r_n$ and $s_n$ are limited. I don't understand why.

Anne
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  • https://math.stackexchange.com/questions/1469162/is-series-sum-n-1-infty-frac-cosnxn-alpha-for-alpha0-conver?rq=1 – For the love of maths May 24 '20 at 15:24
  • In MathJaX there is a specific command for sin/cos that is \sin or \cos. Regards. – Sebastiano May 24 '20 at 15:25
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    Someone (you or the author) is simply applying Dirichlet's test incorrectly. Saying $\sigma_n=\sum_{n=1}^\infty\sin(nx)$ is bounded makes very little sense, and in any case it'ssimply not what's required. You need to show that $s_n=\sum_{j=1}^n\sin(jx)$ is bounded. – David C. Ullrich May 24 '20 at 15:26
  • @DavidC.Ullrich sorry i've edited the text – Anne May 24 '20 at 15:49
  • Your last line of calculation is all wrong. First, $|\sum_{n=1}^\infty e^{ikx}|\ne \sum_{n=1}^\infty \sqrt{r_n^2+s_n^2} $; second, that last sum is obviously $n$, not what you said. For that matter, the preovious $\sum_{n=1}^\infty e^{ikx}= \frac{e^{i(n+1)x}-1}{e^{ix}-1}$ is also wrong; the series on the left obviously diverges. What did you really mean here? – David C. Ullrich May 24 '20 at 16:05
  • Actually it is written in my book and I'm trying to understand – Anne May 24 '20 at 18:45

1 Answers1

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Let me correct some mistakes.

  • You should have $\sum_{k=1}^n e^{ikx} = e^{ix} \frac{e^{inx} - 1}{e^{ix} - 1}$

(There are two errors here: first the sum should not be to infinity, and second, you are using a sum that would work if the summation were for $k$ going from $0$ to $n$ rather than $1$ to $n$.)

  • It is not true that $\sum_{k=1}^n e^{ikx} = \sum_{k = 1}^n \sqrt{r_k^2 + s_k^2}$.

Here you appear to be making the mistake of believing that $\left| \sum_{k = 1}^n z_k \right| = \sum_{k = 1}^n |z_k|$. In fact, we have the triangle inequality $\left| \sum_{k = 1}^n z_k \right| \leq \sum_{k = 1}^n |z_k|$, but not equality.

Using the triangle inequality is key here. You can prove the $|e^{inx} - 1| \leq 2$ part that way.

Anonymous
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