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?
 A: 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 $
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\sin{x} & = 2\sin\pars{x \over 2}\cos\pars{x \over 2} =
2\bracks{2\sin\pars{x \over 4}\cos\pars{x \over 4}}\cos\pars{x \over 2}
\\[5mm] & =
2^{2}\bracks{2\sin\pars{x \over 8}\cos\pars{x \over 8}}
\cos\pars{x \over 4}\cos\pars{x \over 2}
\\[5mm] & = \cdots =
2^{N}\sin\pars{x \over 2^{N}}\prod_{n = 1}^{N}\cos\pars{x \over 2^{N}}
\end{align}

$\ds{\implies}$

\begin{align}
\ln\pars{\sin\pars{x}} & =
N\ln\pars{2} + \ln\pars{\sin\pars{x \over 2^{N}}} +
\sum_{n = 1}^{N}\ln\pars{\cos\pars{x \over 2^{n}}}
\end{align}

Derive both sides respect of $\ds{x}$:

\begin{align}
\cot\pars{x} & =
{1 \over 2^{N}}\cot\pars{x \over 2^{N}} -
\sum_{n = 1}^{N}{1 \over 2^{n}}\tan\pars{x \over 2^{n}}
\end{align}

Takes the limit $\ds{N \to \infty}$ with $\ds{x = 1}$:

$$
\bbx{\sum_{n = 1}^{\infty}{1 \over 2^{n}}\tan\pars{1 \over 2^{n}} = 1 - \cot\pars{1}} \approx 0.3579
$$
