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Find the partial sum of the series and the limit of it:


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?


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 Santos Oct 22 '18 at 21:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


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}


\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 $\ds{x = 1}$:

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


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