## Calculus (3rd Edition)

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.