Prove $\pi = 2\sqrt3 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)3^n}$

Question

We wish to prove the following:

Use the Power Series for $\tan^{-1}(x)$ to show that $$\pi = 2\sqrt 3 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)3^n}$$

We have found that $$\tan^{-1}(x) = \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{2n+1}$$

But we are uncertain of what value of $x$ we may plug in in order to obtain the stated formula. any hints would be greatly appreciated!

Answer 1

It is obvious that $x=\frac1{\sqrt3}$ will be perfect

Answer 2

We have to do two things with our value for $x$: on the one hand we have to get rid of the alternating minus sign and on the other hand we need to get this exponential $\frac1{3^n}$ in somehow. So lets try $-\frac13$. Plugging in gives us $$\tan^{-1}\left(-\frac13\right)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}\left(-\frac13\right)^{2n+1}=-\frac13\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\frac1{3^{\color{red}{2n}}}$$ Close, but not right. Let us try $-\frac1{\sqrt3}$ instead to get rid of the unneeded $2$ within the exponent. Now we obtain $$\tan^{-1}\left(-\frac1{\sqrt3}\right)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}\left(-\frac1{\sqrt3}\right)^{2n+1}=-\frac1{\sqrt3}\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\frac1{3^n}$$ This looks promising! Multiply both sides by $-6$ we get the following $$6\tan^{-1}\left(\frac1{\sqrt3}\right)=2\sqrt3\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\frac1{3^n}$$ All that remains is to reexpress the LHS in terms of $\pi$. Recalling that the tangent is defined as ratio of sine and cosine we may try $\frac\pi6$. Apparently, this works out and we are done.

$$\therefore~2\sqrt3\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\frac1{3^n}~=~\pi$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} 2\root{3}\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}3^{n}} & = -2\root{3}\ic\sum_{n = 0}^{\infty}{\ic^{2n + 1} \over \pars{2n + 1}3^{\pars{2n + 1}/2 - 1/2}} \\[5mm] & = -6\ic\sum_{n = 0}^{\infty}{\pars{\ic/\root{3}}^{2n + 1} \over 2n + 1} \\[5mm] & = -6\ic\sum_{n = 1}^{\infty}{\pars{\root{3}\ic/3}^{n} \over n}\, {1 - \pars{-1}^{n} \over 2} \\[5mm] & = 6\,\Im\sum_{n = 1}^{\infty}{\pars{\root{3}\ic/3}^{n} \over n} \\[5mm] & = -6\,\Im\ln\pars{1 - {\root{3} \over 3}\,\ic} \\[5mm] & = -6\,\arctan\pars{-\root{3}/3 \over 1} = -6\pars{-\,{\pi \over 6}} \\[5mm] & = \bbx{\large \pi} \\ & \end{align}