How to determine the convergence of the following series I was wondering how to determine the convergence of this series:
$$\sum_{n=1}^{+\infty} \left(e^{\frac{1}{4n^2}}\cos \frac{1}{n}\right)^{n^3}$$
I tried to solve it on my own but I have not yet solved it, I think it is possible to use first comparison test and later root criterion to simplify the exponential. Thank you very much for your answers.
 A: The series converges according to the root test. Let $a_n=\left(e^{\frac{1}{4n^2}}\cos\left(\frac 1n\right)\right)^{n^3}$ be the term of the series. Then
$$\begin{align}
\lim_{n\to\infty}\left(a_n\right)^{\frac 1n}&=\lim_{n\to\infty}\left(e^{\frac{1}{4n^2}}\cos\left(\frac 1n\right)\right)^{n^2} \\
&=\lim_{t\to 0^+}\left(e^{\frac{t^2}{4}}\cos(t)\right)^{\frac{1}{t^2}} \\
&=e^{\frac 14}\lim_{t\to 0^+}\left(\cos t\right)^{\frac{1}{t^2}}
\end{align}$$
with the substitution $t=\frac{1}{n}$. We then have
$$\begin{align}
\lim_{t\to 0^+}\left(\cos t\right)^{\frac{1}{t^2}}&=\exp\left(\lim_{t\to 0^+}\frac{\log(\cos t)}{t^2}\right) \\
&=\exp\left(\lim_{t\to 0^+}\frac{\cos t-1}{t^2}\right) =e^{-\frac 12}
\end{align}$$
using $\lim_{y\to 1}\frac{\log y}{y-1}=1$ and $\lim_{y\to 0}\frac{1-\cos y}{y^2}=\frac 12$ (or use L'Hôpital's rule). Thus
$$\lim_{n\to\infty}\left(a_n\right)^{\frac 1n}=e^{\frac 14}e^{-\frac 12}=e^{-\frac 14}<1 $$
A: Another answer, which allows you to bound the sum from above. Every term in the series is positive, so we only need to show that it is bounded by a convergent series. First, note that $\cos(x) \leq 1-x^2/3 $ for all $ 0 \leq x \leq 1 $. Thus, using $ 1-x \leq e^{-x} $, we have that $(\cos(1/n))^{n^3} \leq (1-1/(3n^2))^{n^3} \leq e^{-n/3} $. And so each term in the sum is bounded above by $ (e^{1/(4n^2)}\cos(1/n))^{n^3} \leq e^{n/4-n/3} = e^{-n/12} $. Hence the sum is bounded by the geometric series $ \sum_{n=1}^{\infty}e^{-n/12} = 1/(e^{1/12}-1) $.
