Converge or diverge? : $\sum_{n=1}^{\infty}\frac{\tan{n}}{2^{n}}$

Determine this series converge or diverge, and if it converges, find its value.

$$\sum_{n=1}^{\infty}\frac{\tan{n}}{2^{n}}$$

This was too hard for me, as unboundedness of $\tan{x}$ at infinite number of poles make harder to guess the behavior of value of tangent function at positive integers.

Thanks.

The key, as tetori notes in a comment, is that $\pi$ is known to have a finite irrationality measure: this allows us to state that there's some constant $\theta$ such that there are only finitely many $n$ with $|\pi-\dfrac{m}{n}|\lt \dfrac{1}{n^\theta}$, or in other words $|n\pi-m|\lt n^{1-\theta}$ only finitely often. (For concreteness, we could take $\theta=8$.) Since $\tan(\frac\pi2+\epsilon)=\dfrac{1+\cos(\pi+2\epsilon)}{sin(\pi+2\epsilon)}=\dfrac{\cos(2\epsilon)-1}{\sin(2\epsilon)}$, we can bound the tangent function near a pole from above in absolute value by $\dfrac{2}{\epsilon-\frac{\epsilon^3}{6}} = \epsilon^{-1}\left(\dfrac{2}{1-\frac{\epsilon^2}{6}}\right) = 2\epsilon^{-1}+\dfrac{\epsilon}{3}+\ldots$ — essentially, by $C\epsilon^{-1}$. Now, if $n$ is 'near' a pole of the tangent function, then that means that $n-(m+\frac12)\pi$ is small; or in other words that $2n-(2m+1)\pi$ is small. But we know from the aforementioned result that this quantity can only be smaller than $(2m+1)^{1-\theta}\approx \left(\frac2\pi n\right)^{1-\theta}=\left(\frac2\pi\right)^{1-\theta}n^{1-\theta}$ finitely often. Putting this together with the estimates on $\tan(\frac\pi2+\epsilon)$ we get that there are constants $C$ and $\theta$ such that $|\tan(n)|\gt Cn^{\theta-1}$ only finitely often. Throwing away these exceptions (as they won't affect convergence), we get finally that $\displaystyle\sum_{n=1}^\infty\frac{|\tan n|}{2^n}\leq K+L\cdot \sum_{n=1}^\infty\frac{n^\phi}{2^n}$ for some constants $K, L, \phi$; since the latter series is convergent, then so is $\displaystyle\sum\frac{\tan n}{2^n}$ (and in fact, it's absolutely convergent).