For which real numbers x is the series $$\sum^{\infty}_{n=0} x^n \tan \left(\frac {x}{2^n}\right)$$ convergent and how (i.e. absolutely/conditionally)?
I have proved that the series converges absolutely for $|x|<2$, since $$\left| \frac{a_{n+1}}{a_n} \right|=\frac{|x|}{2} \left|1-\tan ^{2}\frac{x}{2^{n+1}}\right| \rightarrow\frac{|x|}{2}$$ and using the ratio test. Now I need to discuss the part where $|x| \ge 2$.
If $|x|>2 $ then $$\lim_{n\to \infty } x^n \tan \frac{x}{2^n} =\lim_{n\to \infty } 2 (\frac{x}{2})^{n+1}\frac{ \tan \frac{x}{2^n}}{\frac{x}{2^n}} =\infty.$$ Therefore the series diverges