Convergence/divergence of $\sum\limits_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{n^3}$ How do I show convergence/divergence of the series
$$\sum\limits_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{n^3}?$$
I begin by writing $\left(\cos\frac{1}{n}\right)^{n^3} = e^{n^3\ln\left(\cos\frac{1}{n}\right)}$ and continue by Taylor expanding around $0$; first cosine, then ln. But I get nowhere. I would appreciate any help.
 A: You were on the right track.  
First, note that 
$$\cos(1/n)=1-\frac1{2n^2}+O(1/n^4)$$ 
Second, we have
$$\begin{align}
n^3\log(\cos(1/n))&=n^3\log\left(1-\frac1{2n^2}+O(1/n^4)\right)\\\\
&=n^3\left(-\frac1{2n^2}+O(1/n^4)\right)\\\\
&=-\frac12n+O\left(\frac1n\right)
\end{align}$$
Finally, 
$$\begin{align}
e^{n^3\log(\cos(1/n))}&=e^{-\frac12n+O\left(\frac1n\right)}\\\\
&=e^{-n/2}\left(1+O\left(\frac1n\right)\right)
\end{align}$$
Inasmuch as $\sum_{n=1}^\infty e^{-n}$ converges, the series of interest does likewise.
A: Use comparison test, using the inequality mentioned here
$$1-\frac{x^2}{2}\leq \cos{x}\leq e^{-\frac{x^2}{2}}, x \in \left[0,\frac{\pi}{2}\right]$$
or for $n\geq1$
$$0<\cos{\frac{1}{n}}\leq e^{-\frac{1}{2n^2}}$$
thus 
$$0<\left(\cos{\frac{1}{n}}\right)^{n^3}\leq e^{-\frac{n}{2}}=\left(\frac{1}{\sqrt{e}}\right)^n$$
and $0<\frac{1}{\sqrt{e}}<1$. Finally
$$0<\sum\limits_{n=1}\left(\cos{\frac{1}{n}}\right)^{n^3}\leq \sum\limits_{n=1}\left(\frac{1}{\sqrt{e}}\right)^n=\frac{1}{\sqrt{e}}\cdot\left(\frac{1}{1-\frac{1}{\sqrt{e}}}\right)=\frac{1}{\sqrt{e}-1}$$
A: $\cos(\frac{1}{n}) = 1 - \frac{1}{2n^2} + o(\frac{1}{n^3})$. Let $a_n = \cos(\frac{1}{n})$ and $b_n = 1- \frac{1}{2n^2}$. We have $$\frac{a_n^{n^3}}{b_n^{n^3}} = (1 + \frac{o(\frac{1}{n^3})}{1-\frac{1}{2n^2}})^{n^3} = (1+c_n o(1))^{n^3} $$ where $c_n = \frac{1}{n^3} \cdot \frac{1}{1 - \frac{1}{2n^2}} = \frac{1}{n^3-\frac{n}{2}} $, so since $c_n \cdot n^3 \to 1$ we get $\frac{a_n^{n^3}}{b_n^{n^3}} \to 1$. So by asymptotics (note that $a_n,b_n$ are nonnetative for large $n$), the question is equivalent to convergence/divergence of $\sum_{n=1}^\infty (1-\frac{1}{2n^2})^{n^3}$. Now, taking $n'$th root we get $\exp(n^2\ln(1-\frac{1}{2n^2}))$. To find its limit, note that $\exp$ is continuous, so it is sufficient to find the limit of sequence $(n^2\ln(1-\frac{1}{2n^2}))$ which is equal to $-\frac{1}{2} \cdot \frac{\ln(1-\frac{1}{2n^2})}{-\frac{1}{2n^2}} \to -\frac{1}{2}$, so our limit tends to $\exp(-\frac{1}{2}) < 1$, and that means our series converges.
A: A possible way to show convergence is to rewrite 
$$\cos \frac 1n = \cos \frac{2}{2n} = 1- 2\sin^2 \frac 1{2n} $$
and now use root test and the standard limits $\lim_{t\to 0}(1-t)^{\frac 1t} = \frac 1e$ and $\lim_{t\to 0}\frac{\sin t}{t}=1$:
\begin{eqnarray*}\sqrt[n]{\left(1- 2\sin^2 \frac 1{2n}\right)^{n^3}}
& = & \left(\left(1- 2\sin^2 \frac 1{2n}\right)^{\frac{1}{2\sin^2 \frac 1{2n}}}\right)^{n^2\cdot 2\sin^2 \frac 1{2n}}\\
& \stackrel{n\to \infty}{\longrightarrow} & \left(\frac 1e\right)^{\frac 12}=\frac 1{\sqrt e} < 1
\end{eqnarray*}
A: The ratio test works fine using your way
$$a_n=\cos ^{n^3}\left(\frac{1}{n}\right)\implies \log(a_n)=n^3\log\left(\cos \left(\frac{1}{n}\right) \right)$$
$$\cos \left(\frac{1}{n}\right)=1-\frac{1}{2 n^2}+\frac{1}{24 n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log\left(\cos \left(\frac{1}{n}\right) \right)=-\frac{1}{2 n^2}-\frac{1}{12 n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log(a_n)=n^3\left(-\frac{1}{2 n^2}-\frac{1}{12 n^4}+O\left(\frac{1}{n^6}\right) \right)=-\frac{n}{2}-\frac{1}{12 n}+O\left(\frac{1}{n^3}\right)$$ Now, apply twice and continue with Taylor series
$$\log(a_{n+1})-\log(a_n)=-\frac{1}{2}+\frac{1}{12 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{a_{n+1}}{a_{n}}=e^{\log(a_{n+1})-\log(a_n) }=\frac{1}{\sqrt{e}}\left(1+\frac{1}{12   n^2}\right)+O\left(\frac{1}{n^3}\right)$$
