is my argument for $\lim_{n\rightarrow\infty}(a_n) = 0$ given $a_n = \tan(n) (\frac{1}{e})^{n}$ correct? I want to show that $\lim_{n\rightarrow\infty}(a_n) = 0$ where
$a_n = \tan(n) (\frac{1}{e})^{n}$
My argument is that:
$\tan(n) = \frac{\sin(n)}{\cos(n)} \in \mathbb{R}$ for $n \in \mathbb{N}$ since cos(x) only has roots of the form $x = -\frac{\pi}{2} + \pi k , k \in \mathbb{Z}$.
The convergence of the other term is obvious.
Is it correct?
 A: @Daniel Fischer already pointed out its relevance to the notion of irrationality measure.

This is related to the following question:

How fast $n$ gets closer to the zero set $\frac{\pi}{2} + \pi \Bbb{Z}$ of $\cos x = 0$?

We do not want a situation where (a subsequence of) $\cos n$ decays so fast that it can even beat the exponential factor $e^{-n}$. In other words, we want that $n$ stays moderately far from the zero set $\frac{\pi}{2} + \pi \Bbb{Z}$.
Related to this question is the irrationality measure of $1/\pi$. In particular, if the irrationality measure $\mu$ of $1/\pi$ is finite, then for each $\epsilon > 0$ there exists $c = c(\epsilon) > 0$ such that
$$ \forall q \in \Bbb{N}^{+}, p \in \Bbb{Z}, \quad \left| \frac{1}{\pi} - \frac{p}{q} \right| \geq \frac{c}{q^{\mu+\epsilon}}. $$
Now let us plug $q = 2n$ and $p = 2k+1$. Manipulating the inequality a little bit, we find that for some constant $c' = c'(\epsilon) > 0$,
$$ |n - (k+\tfrac{1}{2})\pi| \geq c' n^{-(\mu+\epsilon-1)}. $$
This implies that $\cos n$ stays away from $0$ in a predictable way: if $n$ is large, then
$$ |\cos n| \geq |\sin(c' n^{-(\mu+\epsilon-1)})| \geq \frac{2c'}{\pi} n^{-(\mu+\epsilon-1)}. $$
So it follows that
$$ |e^{-n}\tan n| \leq C n^{\mu+\epsilon-1} e^{-n} \xrightarrow[n\to\infty]{ } 0. $$
Finally, it is proven that $\mu$ is indeed finite. Therefore $e^{-n} \tan n$ converges to $0$.
