A calculus problem by Syed Hamza Khalid

Recall that $|x|\leq 1$ $\begin{aligned} & \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + \cdots \\& \ln(1-x) = -x - \frac{x^2}{2}-\frac{x^3}{3} - \cdots \end{aligned}$

Subtracting the above series as $\begin{aligned} & \ln(1+x)-\ln(1-x) = 2x +\frac{2x^3}{3} + \frac{2x^5}{5}+\cdots \\& \ln\left(\frac{1+x}{1-x}\right) = 2\left(x + \frac{x^3}{3}+\frac{x^5}{5} +\cdots \right) \\& \ln\left(\frac{1+ i }{1- i }\right) = 2\left(i +\frac{i^3}{3}+\frac{i ^5}{5}+\cdots \right)\phantom{cccccc} \text{put x = i, where}\text{ i} = \sqrt{-1}\\& \ln(\text{exp}(i \frac{\pi}{2})) = 2i\left(1-\frac{1}{3}+\frac{1}{5}+\cdots\right) \phantom{cccccc}\text{exp(x)}=e^x \\& \frac{\pi}{4} = 1-\frac{1}{3}+\frac{1}{5}\cdots \end{aligned}$

Hence the value of $x=\pi$ .