$ \sum_{k=1}^{+\infty} \frac{\sin^2 (kx)}{k}$ and $ \sum_{k=1}^{+\infty} \frac{\cos^2 (kx)}{k}$

Question

Let's examine the series $ \sum\limits_{k=1}^{+\infty} \dfrac{\sin^2 (kx)}{k}$ and $ \sum\limits_{k=1}^{+\infty} \dfrac{\cos^2 (kx)}{k}$

My attempt :
$\forall t , ~ \cos^2(t) + \sin^2(t) =1$ and $\sum\limits_{k \ge 1} \dfrac{1}{k} =\infty$ .
As the two terms are positives, at least one of the series should be divergent.
How to prove that both series are divergent ?

As given in hint, $\cos^2(kx)= 1 + 2 \cos(2kx)$

Mark Viola · Accepted Answer · 2020-08-18T15:39:01.870

1

HINT:

Both series diverge. To show this, make use of the identities

$$\begin{align} \sin^2(x)&=\frac{1-\cos(2x)}{2}\\\\ \cos^2(x)&=\frac{1+\cos(2x)}{2} \end{align}$$

along with the fact that $\sum_{n=1}^\infty \frac{\cos(2nx)}{n}$ converges for $x\ne m\pi$, $m\in \mathbb{Z}$, as guaranteed by Dirichlet's test.

edited Aug 18 '20 at 15:39

answered Aug 18 '20 at 15:33

Mark Viola

179,405

The diference has a general term $\frac{cos(kx)}{k}$ that diverges. The sum is a serie that diverges. – zestiria Aug 18 '20 at 15:44
Note that $$\sum_{n=1}^N \frac{\cos^2(nx)}{n}=\sum_{n=1}^N\left(\frac{1+\cos(2nx)}{n}\right)=\sum_{n=1}^N \frac1n +\frac12\sum_{n=1}^N\frac{\cos(2nx)}{n}$$The harmonic series diverges, but $\sum_{n=1}^N \frac{\cos(2nx)}{n}$ converges when $x\ne m \pi$. What does that tell you about the original series? – Mark Viola Aug 18 '20 at 15:47
The sum of one serie converges and one serie diverges, diverges. So the original serie diverges. – zestiria Aug 18 '20 at 15:56
Yes, indeed. You have it now. Well done. – Mark Viola Aug 18 '20 at 16:00
Thanks, I will try to do the Dirichlet's test. – zestiria Aug 18 '20 at 16:04
You're welcome. My pleasure. – Mark Viola Aug 18 '20 at 16:47

$ \sum_{k=1}^{+\infty} \frac{\sin^2 (kx)}{k}$ and $ \sum_{k=1}^{+\infty} \frac{\cos^2 (kx)}{k}$

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