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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.

J.Dane
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  • What does $n3$ mean? It makes a huge difference whether you mean $3n$ or $n^3$. – B. Goddard Jul 25 '18 at 13:52
  • it's $n^3$ . I edited it – J.Dane Jul 25 '18 at 13:56
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    I don't think there are "so many" theorems. When you have an $n$th power (or worse) then the root test suggests itself. – B. Goddard Jul 25 '18 at 13:57
  • I said I'm new to this. I am trying to learn which theorem should I use for specific problems – J.Dane Jul 25 '18 at 13:59
  • @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! – user9750060 Jul 25 '18 at 14:15
  • Yes, thank you very much! – J.Dane Jul 25 '18 at 14:17
  • And I told you which theorem to use and why. Who knows why you'd have a problem with that. – B. Goddard Jul 25 '18 at 15:15

4 Answers4

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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

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

user
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    I don’t understand the downvote. – Szeto Jul 25 '18 at 14:09
  • @Szeto Thanks for your opinion, I was wondering the same and looking for the mistake! So you agree with that way to solve? – user Jul 25 '18 at 14:11
  • Maybe it got downvoted, because the solution is not elementary enough? – Cornman Jul 25 '18 at 14:21
  • 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... – David C. Ullrich Jul 25 '18 at 14:25
  • @DavidC.Ullrich On the other han some kind of elementary questions very poorly posed are ignored, as for example (and only as an example) https://math.stackexchange.com/q/2862389/505767 – user Jul 25 '18 at 14:29
  • I wasn't saying it was right, or consistent, just pointing out a possible reason for the downvote. Heh, there's a difference of opinion regarding the limit in the root test - maybe someone thought yours was wrong (it's right). – David C. Ullrich Jul 25 '18 at 14:36
  • @MarkViola I need to learn from you the way to use inequalities for series, you know that I'm "LCT" addicted :P – user Jul 25 '18 at 14:59
  • Limit comparison test :) – user Jul 25 '18 at 15:01
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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 $2$: $$\cos u=1-\frac{u^2}2+o(u^2).$$

Bernard
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    I think it should be $n^2$ not $1/n^2$ – J.Dane Jul 25 '18 at 14:30
  • @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! – Bernard Jul 25 '18 at 14:39
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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.

Mark Viola
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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 $$0\leq \sum\limits_{n=1} \left(\cos{\frac{1}{\sqrt{n}}}\right)^{n^3}\leq \sum\limits_{n=1}e^{-\frac{n^2}{2}}$$

rtybase
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