<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] 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
$$\log(a_{n+1})-\log(a_n)=-\frac{1}{2}+\frac{1}{12 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{a_{n+1}}{a_{n}}=e^{\log(a_{n+1})-\log(a_n) }=\frac{1}{\sqrt{e}}\left(1+\frac{1}{12   n^2}\right)+O\left(\frac{1}{n^3}\right)$$
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