Check convergence of the series $\sum_{n=1}^\infty\cos^{n^3}\frac1{\sqrt n}$

Question

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.

Answer 1

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 $0<x\le \pi/2$).

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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.

Answer 2

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

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

Answer 3

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 $2$: $$\cos u=1-\frac{u^2}2+o(u^2).$$

Answer 4

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