This is my functional series: $$\sum_{k=1}^\infty x^k \tan(\frac{x}{2^k}) $$
Now, to get the convergence radius (not sure if that's the correct word for it in English), I've been taught that I need to take the absolute value of $a_k$ and solve it. So I tried using the ratio test and got this:
$ \lim \limits_{k \to \infty} \frac{|x \tan(\frac{x}{2^{k+1}})|}{|\tan(\frac{x}{2^{k}})|}$
WolframAlpha tells me that the answer to this is $\frac{|x|}{2}$. I have no idea how to get it, could somebody help me, is there some kind of a formula we can use for tangents?
After that we should get that the convergence radius/area is |x| < 2 (because convergence area for ratio test is < 1).
But after that I have been told that I should double-check manually |x| = 2 and |x| > 2.
Wolfram tells me that $ \lim \limits_{k \to \infty} |2^k \tan(\frac{2}{2^{k}})| = 2$ and therefore x=2 diverges. But again I have no idea how to actually get it.
The limit for $ \lim \limits_{k \to \infty} |(-2)^k \tan(\frac{-2}{2^{k}})|$ doesn't exist so that diverges as well.
Showing |x| > 2 is also quite confusing for me.