<s>[INST] 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.
 [/INST] $\cos(\frac{1}{n}) = 1 - \frac{1}{2n^2} + o(\frac{1}{n^3})$. 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.
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