4
$\begingroup$

I'm trying to figure out if the following series converges or diverges. I have spent hours on it and can't figure it out. Tried to use the comparison test, Dirichlet, and Abel, but none of it worked. Any hints?

$$ \sum_{n=1}^\infty \frac{\cos^2(n)}{\sqrt{n}} $$

$\endgroup$
  • 1
    $\begingroup$ Hey can you show what you did for the Abel comparasin test? $\endgroup$ – Max0815 Feb 6 at 19:38
8
$\begingroup$

HINT:

Recall that $\cos^2(x)=\frac{1+\cos(2x)}{2}$.

$\endgroup$
  • $\begingroup$ Is using this and splitting the series into two helpful? Is the addition of a divergent and a convergent series also divergent? $\endgroup$ – peroxisome7 Feb 6 at 19:57
  • $\begingroup$ @peroxisome7, Yes. That is a version of contrapositive statement of the fact that difference of two convergent series is again convergent. $\endgroup$ – Sangchul Lee Feb 6 at 20:07
  • 1
    $\begingroup$ @peroxisome Note that $$\sum_{n=1}^N \frac{\cos^2(n)}{\sqrt n} =\frac12 \sum_{N=1}^N \frac{1}{\sqrt n}+\frac12 \sum_{n=1}^N \frac{\cos(2n)}{\sqrt n}$$Now, Dirichelt's test shows that the second partial sum on the right-hand side converges. If the partial sum on the left-hand side converged, then the first partial sum on the right-hand side would be the sum (difference) of two convergent partial sums and hence would also converge. Inasmuch as the first partial sum on the right-hand side diverges, what can one conclude? $\endgroup$ – Mark Viola Feb 6 at 21:20
8
$\begingroup$

Another approach is to observe that among $e^{in}, n=1,2,\dots, 7,$ at least one of these points lies in the arc $\{e^{it}: t\in (-\pi/4,\pi/4)\}.$ Thus

$$\sum_{n=1}^{7} \frac{\cos^2 n}{\sqrt n} \ge \frac{1}{ 2}\frac{1}{\sqrt 7}.$$

The same thing happens for $n=8,\dots,14.$ etc. So the series in question is at least

$$\sum_{m=1}^{\infty} \frac{1}{ 2}\frac{1}{\sqrt {7m}} = \infty.$$

Note that this idea will work if $\cos^2 n$ is replaced by $|\cos n|^p$ for any $p>0$.

$\endgroup$
  • $\begingroup$ (+1) I remember having used this trick to solve a previous question on the site. Under its apparence of simplicity, it is indeed rather powerful. $\endgroup$ – Did Feb 6 at 20:05
  • 2
    $\begingroup$ @zhw. Inasmuch as at least one of the points $1,2,\dots, 7$ lies in that arc, $\cos^2(n)\ge 1/2$ for such a point. Then, $\sum_{n=1}^7 \frac{\cos^2(n)}{\sqrt n}\ge \frac1{2\sqrt 7}$, would it not? $\endgroup$ – Mark Viola Feb 6 at 20:42
  • $\begingroup$ @MarkViola Right you are. Now edited. $\endgroup$ – zhw. Feb 6 at 23:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.