does $ \sum _{n=0}^{\infty }\left(\cos^n\left(\frac{1}{\sqrt{n}}\right)-\frac{1}{\sqrt{e}}\right) $ converge? I'm trying to find out whether $\sum _{n=0}^{\infty }\left(\cos^n\left(\frac{1}{\sqrt{n}}\right)-\frac{1}{\sqrt{e}}\right)$ converges or not. I've tried with taylor series but it doesn't lead me anywhere except with the fact that $\lim_{n \to \infty}\cos^n\left(\frac{1}{\sqrt{n}}\right)-\frac{1}{\sqrt{e}}=0$ and therefore it has "a chance" to converge.
Any hint?
 A: The $n$-th term is asymptotic to $$-\frac{1}{12 \sqrt{e} n}$$ so, alas, the series diverges.
A: hint
$$\cos(\frac{1}{\sqrt{n}})=1-\frac{1}{2n}+\frac{1}{24n^2}+o(\frac{1}{n^2})$$
$$\ln(1+\cos(\frac{1}{\sqrt{n}})-1)=$$
$$-\frac{1}{2n}-\frac{1}{12n^2}+o(1/n^2)$$
thus
$$\cos^n(\frac{1}{\sqrt{n}})=e^{n\ln(\cos(\frac{1}{\sqrt{n}}))}$$
$$=e^{-\frac 12}e^{-\frac 1n(\frac{1}{12}+o(1))}$$
$$=\frac{1}{\sqrt{e}}(1-\frac{1}{12n}(1+o(1))$$
A: Note that
$$
\cos\frac{1}{\sqrt{n}}\sim1-\frac{1}{2n}+\frac{1}{24n^2}+\ldots,
$$
so
$$
\log\cos\frac{1}{\sqrt{n}}= \log\left(1-\left(\frac{1}{2n}-\frac{1}{24n^2}+\ldots\right)\right)\\=-\left(\frac{1}{2n}-\frac{1}{24n^2}+\ldots\right)-\frac{1}{2}\left(\frac{1}{2n}+\ldots\right)^2+\ldots\\=-\frac{1}{2n}-\frac{1}{12n^2}+\ldots.
$$
So
$$
\cos^n\frac{1}{\sqrt{n}}=\exp\left(n\log \cos\frac{1}{\sqrt{n}}\right) \\ = \exp\left(-\frac{1}{2}-\frac{1}{12n}+\ldots\right) \\ =\frac{1}{\sqrt{e}}\exp\left(-\frac{1}{12n}+\ldots\right)\\ =\frac{1}{\sqrt{e}}-\frac{1}{12n\sqrt{e}}+O(n^{-2}).
$$
Hence the terms are asymptotic to $-\frac{1}{12n\sqrt{e}}$, and the sum diverges.
