Answer
converges
Work Step by Step
Given $$\sum_{n=1}^{\infty}\left(\cos \frac{1}{n}\right)^{n^{3}}$$
Since $a_n = \left(\cos \dfrac{1}{n}\right)^{n^{3}}$, then by using the root test
\begin{align*}
\rho &= \lim_{n\to \infty } \sqrt[n]{a_n}\\
&= \lim_{n\to \infty } \sqrt[n]{\left(\cos \dfrac{1}{n}\right)^{n^{3}}}\\
&= \lim_{n\to \infty } \left(\cos \dfrac{1}{n}\right)^{n^ {2}}\\
&=\lim _{n \rightarrow \infty}\left(e^{\left.n^{2} \ln \left(\cos \left(\frac{1}{n}\right)\right)\right)}\right)
\end{align*}
By using L'Hopital's rule, we get
\begin{align*}
\lim _{n\to \infty \:}\:n^2\ln \left(\cos \left(\frac{1}{n}\right)\right)&= \lim _{n\to \infty \:}\left(\frac{\ln \left(\cos \left(\frac{1}{n}\right)\right)}{\frac{1}{n^2}}\right)\\
&= \lim _{n\to \infty \:}\left(\frac{\frac{\tan \left(\frac{1}{n}\right)}{n^2}}{-\frac{2}{n^3}}\right)\\
&= -\frac{1}{2}\cdot \lim _{n\to \infty \:}\left(n\tan \left(\frac{1}{n}\right)\right)\\
&=\frac{-1}{2}
\end{align*}
Then $$\rho = e^{-1/2}<1 $$
Hence, the given series converges.
