showing $a_n = \frac{\tan(1)}{2^1} + \frac{\tan(2)}{2^2} + \dots + \frac{\tan(n)}{2^n}$ is not Cauchy

Question

My gut telling me that the following sequence is not Cauchy, but I don't know how to show that. $$a_n = \frac{\tan(1)}{2^1} + \frac{\tan(2)}{2^2} + \dots + \frac{\tan(n)}{2^n}$$

Answer 1

In 2008, Shalikhov has improved the upper bound of irrationality measure of $\pi$ to about 7.6063. This means for any $\mu > 7.6063$, there are at most finitely many pairs of relative prime integers $(p,q)$ such that

$$|\pi - \frac{p}{q}| < \frac{1}{q^{\mu}}$$

A consequence of this is if one choose a small enough $C_\mu > 0$, then for any pair of positive integers $(p,q)$, we have

$$|\pi - \frac{p}{q}| > \frac{C_\mu}{q^{\mu}}$$

In order for $\tan n$ to blow up, $n$ need to be very close to some half integer multiple , $(\ell + \frac12)\pi$, of $\pi$. Using the bound of irrational measure above and let $\mu = 8$, we find

$$\left|n - (\ell + \frac12)\pi\right| = \frac{2\ell+1}{2}\left| \frac{2n}{2\ell+1} - \pi \right| > \frac{C_8}{2(2\ell+1)^7} \sim \frac{\pi^7C_8}{2^8n^7} $$ Notice $$|\tan x| \sim \frac{1}{|x - (\ell+\frac12)\pi|}\quad\text{ for } x \sim (\ell+\frac12)\pi $$ This gives us an approximate bound for $\tan n$

$$|\tan n| \,\lesssim\,\frac{2^8 n^7}{\pi^7C_8} \quad\implies\quad \sum_{n=1}^{\infty} \frac{\tan n}{2^n}\quad\text{ converges absolutely.}$$