<s>[INST] Find the partial sum of the series and the limit of it $\sum_{n=1}^{\infty}\frac{1}{2^n}\tan\frac{1}{2^n}$ Find the partial sum of the series and the limit of it:
$\sum_{n=1}^{\infty}\frac{1}{2^n}\tan\frac{1}{2^n}$
I did $\lim_{n\to\infty}\frac{1}{2^n}\tan\frac{1}{2^n} = 0$ so we can not say that the series is divergent.
I tried to use telescopic sum but I do not know what to do with that $\tan$. Can you give me any hint how to write it?
UPDATE
So I did what @lab bhattacharjee recommended and I got that: $\frac{1}{2}\tan x = \frac{1}{\tfrac{1}{\tan\frac{x}{2}}-\tan\frac{x}{2}}\iff\tan x = \frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}=\frac{1+\tan\frac{x}{2}+\tan\frac{x}{2} - 1}{(1+\tan\frac{x}{2})(1-\tan\frac{x}{2})} = \frac{1}{1-\tan\frac{x}{2}}-\frac{1}{1+\tan\frac{x}{2}}.$
What should I do next?
 [/INST] Hint :
$$\cot x-\tan x =2\cot 2x\iff\dfrac12\tan x=? $$
Put $x=1/2^n, n=1,2,3,\cdots  m$ and add to find  the partial  sum 
Finally set $m\to\infty $
</s>