3
$\begingroup$

Prove the convergence of the series $$\sum_{n=1}^\infty\cos^{n^3}\frac1{\sqrt n}$$ This is the first time that I'm learning about the convergence of the series and there are so many theorems about how to prove one and I really don't know which one to use.

I would really appreciate some help.

$\endgroup$

closed as off-topic by Did, Nosrati, Namaste, Xander Henderson, Parcly Taxel Jul 26 '18 at 1:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Nosrati, Namaste, Xander Henderson, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What does $n3$ mean? It makes a huge difference whether you mean $3n$ or $n^3$. $\endgroup$ – B. Goddard Jul 25 '18 at 13:52
  • $\begingroup$ it's $n^3$ . I edited it $\endgroup$ – J.Dane Jul 25 '18 at 13:56
  • 1
    $\begingroup$ I don't think there are "so many" theorems. When you have an $n$th power (or worse) then the root test suggests itself. $\endgroup$ – B. Goddard Jul 25 '18 at 13:57
  • $\begingroup$ I said I'm new to this. I am trying to learn which theorem should I use for specific problems $\endgroup$ – J.Dane Jul 25 '18 at 13:59
  • $\begingroup$ @J.Dane When proving convergence of series, one of the best tools in my opinion is using the comparison test. There are a bunch of series (geometric, harmonic, alternating harmonic, etc) which have already been proven to converge or diverge. If you are looking to prove convergence for a series A, you can compare it to a series B where every term in B is larger than the the terms in A. If B converges, then A must also converge. Hope this helps! $\endgroup$ – user9750060 Jul 25 '18 at 14:15
1
$\begingroup$

We need neither the root test nor Taylor's Theorem to proceed. Here instead, we use the elementary inequalities $\log(1-x)\le -x$ and $\sin(x)\ge 2x/\pi$ (for $0<x\le \pi/2$).


Proceeding, we see that for $n\ge1$

$$\begin{align} 0\le \cos^{n^3}(n^{-1/2})&=e^{n^3\log(1-2\sin^2(n^{-1/2}/2))}\\\\ &\le e^{-2n^3\sin^2(n^{-1/2}/2)}\\\\ &\le e^{-2n^2/\pi^2}\\\\ \end{align}$$

Inasmuch as $\sum_{n=1}^\infty e^{-2n^2/\pi^2}$ converges, we conclude that the series of interest converges also.

$\endgroup$
  • $\begingroup$ Very nice and elegant solution! $\endgroup$ – gimusi Jul 25 '18 at 14:59
  • $\begingroup$ @gimusi Thank you! Much appreciated. $\endgroup$ – Mark Viola Jul 25 '18 at 15:00
  • $\begingroup$ Would the down voter care to comment???' $\endgroup$ – Mark Viola Jul 26 '18 at 21:11
2
$\begingroup$

We have that

$$\left(\cos\left(\frac{1}{\sqrt n}\right)\right)^{n^3}= \left(1-\frac1{2n}+\frac1{24n^2}+O\left(\frac1{n^3}\right)\right)^{n^3}=e^{n^3\log{\left(1-\frac1{2n}+\frac1{24n^2}+O\left(\frac1{n^3}\right)\right)}}\sim e^{-\frac{n^2}2}$$

therefore the given series converges by limit comparison test with $\sum e^{-\frac{n^2}2}$.

As an alternative by root test for $a_n=\left(\cos\left(\frac{1}{\sqrt n}\right)\right)^{n^3}$ we have

$$\sqrt[n]{a_n}=\left(\cos\left(\frac{1}{\sqrt n}\right)\right)^{n^2}\sim e^{-\frac{n}2}\to 0$$

$\endgroup$
  • 1
    $\begingroup$ I don’t understand the downvote. $\endgroup$ – Szeto Jul 25 '18 at 14:09
  • $\begingroup$ @Szeto Thanks for your opinion, I was wondering the same and looking for the mistake! So you agree with that way to solve? $\endgroup$ – gimusi Jul 25 '18 at 14:11
  • $\begingroup$ Maybe it got downvoted, because the solution is not elementary enough? $\endgroup$ – Cornman Jul 25 '18 at 14:21
  • $\begingroup$ Note there are two close votes on the question (for a totally silly reason, imo). It can happen that an answer gets downvoted simply bbecause people disapprove of answering "bad" questions... $\endgroup$ – David C. Ullrich Jul 25 '18 at 14:25
  • $\begingroup$ @DavidC.Ullrich On the other han some kind of elementary questions very poorly posed are ignored, as for example (and only as an example) math.stackexchange.com/q/2862389/505767 $\endgroup$ – gimusi Jul 25 '18 at 14:29
2
$\begingroup$

The simplest here is to use the root test: you should find the $$\lim_{n\to\infty}\biggl(\cos^{n^3}\!\frac1{\sqrt n}\biggr)^{\!\tfrac1n}=\lim_{n\to\infty}\cos^{n^2}\!\frac1{\sqrt n}=0.$$

Hint:

This is equivalent to showing $\;\lim_{n\to\infty}n^2\log\biggl(\cos\dfrac1{\sqrt n}\biggr)=-\infty$, and you can use for that Taylor formula at order $2$: $$\cos u=1-\frac{u^2}2+o(u^2).$$

$\endgroup$
  • 1
    $\begingroup$ I think it should be $n^2$ not $1/n^2$ $\endgroup$ – J.Dane Jul 25 '18 at 14:30
  • $\begingroup$ @DavidC.Ullrich: You're right. I should always check computations with a pencil and paper before posting, but I didn't. I've fixed the answer. Thank you for pointing the error! $\endgroup$ – Bernard Jul 25 '18 at 14:39
0
$\begingroup$

In style to @MarkViola's version. Here is a proof of $$0\leq\cos{x}\leq e^{-\frac{x^2}{2}}, \forall x\in \left[0,\frac{\pi}{2}\right]$$ using nothing but derivatives, so may be classified as elementary. Also, $0<\frac{1}{\sqrt{n}}<\frac{\pi}{2}, \forall n >0$. Then $$0\leq\cos{\frac{1}{\sqrt{n}}}\leq e^{-\frac{1}{2n}}\Rightarrow 0\leq \left(\cos{\frac{1}{\sqrt{n}}}\right)^{n^3}\leq e^{-\frac{n^2}{2}} \tag{1}$$ and finally $$0\leq \sum\limits_{n=1} \left(\cos{\frac{1}{\sqrt{n}}}\right)^{n^3}\leq \sum\limits_{n=1}e^{-\frac{n^2}{2}}$$

$\endgroup$

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