Check convergence of the series $\sum_{n=1}^\infty\cos^{n^3}\frac1{\sqrt n}$ 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.
 A: 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
$$\sqrt[n]{a_n}=\left(\cos\left(\frac{1}{\sqrt n}\right)\right)^{n^2}\sim e^{-\frac{n}2}\to 0$$
A: 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 $2$:
$$\cos u=1-\frac{u^2}2+o(u^2).$$
A: 
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.
A: 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
$$0\leq \sum\limits_{n=1} \left(\cos{\frac{1}{\sqrt{n}}}\right)^{n^3}\leq \sum\limits_{n=1}e^{-\frac{n^2}{2}}$$
