Infinite Circles

Let $S = 4 - \dfrac 43 + \dfrac 45 - \dfrac 47 + \dfrac 49 + \cdots$ . We note that

$\begin{aligned} \ln \left(\frac {1+x}{1-x} \right) & = 2x + \frac 23 x^3 + \frac 25x^5 + \frac 27x^7 + \frac 29 x^9 + \cdots \\ \ln \left(\frac {1+i}{1-i} \right) & = 2 \left(i - \frac i3 + \frac i5 - \frac i7 + \frac i9 + \cdots \right) \\ 2 \ln \left(\frac {1+i}{1-i} \right) & = i\left(4 - \frac 43 + \frac 45 - \frac 47 + \frac 49 + \cdots \right) \\ \implies iS & = 2 \ln \left(\frac {1+i}{1-i} \right) \\ & = 2 \ln \left(\frac {(1+i)^2}{(1-i)(1+i)} \right) \\ & = 2 \ln \left(\frac {2i}2 \right) = 2 \ln \color{#3D99F6} i & \small \color{#3D99F6} \text{By Euler's formula: }e^{ix} = \cos x + i\sin x \\ & = 2 \ln {\color{#3D99F6} e^{i \frac \pi 2}} = 2 \cdot i \frac \pi 2 = i \pi \\ \implies S & = \pi \approx \boxed{3.142} \end{aligned}$

Use the sum $\frac { \pi }{ 4 } =1-\frac { 1 }{ 3 } +\frac { 1 }{ 5 } -\frac { 1 }{ 7 } +\frac { 1 }{ 9 } - ...$ then multiply both sides by 4 to get $\pi =4(1-\frac { 1 }{ 3 } +\frac { 1 }{ 5 } -\frac { 1 }{ 7 } +\frac { 1 }{ 9 } - ...)$ then distribute the four to get $\pi =4-\frac { 4 }{ 3 } +\frac { 4 }{ 5 } -\frac { 4 }{ 7 } +\frac { 4 }{ 9 } -...$