Find the partial sum of the series and the limit of it $\sum_{n=1}^{\infty}\frac{1}{2^n}\tan\frac{1}{2^n}$ [duplicate]

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?

marked as duplicate by Hans Lundmark, Cesareo, ArsenBerk, rtybase, José Carlos SantosOct 22 '18 at 21:59

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 $$m\to\infty$$

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \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 $$\ds{x = 1}$$:

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