I need to confirm the Leibniz series through using Fourier series. I have done what I believe is the correct Fourier series just want to know if I am on the right path? If it is correct I don't see any way to evaluate it that comes out as Leibniz.
Leibniz: \begin{equation} \frac{\pi}{4}=\sum_{i=0}^\infty \frac{(-1)^i}{(2i+1)} \end{equation}
Function: $$f(x) = \begin{cases} 0, &\text{for}\ -\pi<x<0 \\ 1, &\text{for}\ 0\leq x<\pi \end{cases} $$
Fourier: \begin{equation} b_n = \frac{1}{\pi}\int_{-\pi}^\pi f(x) \sin(xn) dx=\frac{1}{\pi}(\int_{0}^\pi 1 \sin(xn) dx+\int_{-\pi}^0 0 \sin(xn) dx)=\frac{1}{\pi}(1/n)(-1(-1)^n+1)+0)=\frac{1-(-1)^n}{\pi*n} \end{equation} \begin{equation} a_n =\frac{1}{\pi}(\int_{0}^\pi 1 \cos(xn) dx +\int_{-\pi}^0 0 \cos(xn) dx) = \frac{1}{\pi}(0+0)=0 \end{equation} \begin{equation} a_0 = \frac{1}{\pi}(\int_{0}^\pi 1 dx +\int_{-\pi}^0 0 dx) = \frac{1}{\pi}(\pi-0+0)=1 \end{equation}
\begin{equation} f(x) = \frac{a_0}{2}+\sum_{i=1}^\infty (a_n\cos(nx)+b_n\sin(nx) =\frac{1}{2}+\sum_{i=1}^\infty (\frac{1-(-1)^n}{\pi*n}\sin(nx) \end{equation}
What you have obtained is that
$$ f(x) = \frac{1}{2} + \sum_{n=1}^{\infty}\frac{1 - (-1)^n}{\pi n} \sin(nx). $$ Therefore, it follows that, for $x = \frac{\pi}{2}$, $$ \begin{align} f(\pi/2) = 1 &= \frac{1}{2} + \sum_{n=1}^{\infty}\frac{1 - (-1)^n}{\pi n} \sin(n\pi/2) \\ \text{using the note from below} \\ &= \frac{1}{2} + \sum_{k=1}^{\infty}\frac{1 - (-1)^{4k-3}}{\pi (4k-3)}\sin\left(\frac{\pi(4k-3)}{2}\right) + \sum_{k=1}^{\infty}\frac{1 - (-1)^{4k-2}}{\pi (4k-2)}\sin\left(\frac{\pi(4k-2)}{2}\right) + \sum_{k=1}^{\infty}\frac{1 - (-1)^{4k-1}}{\pi (4k-1)}\sin\left(\frac{\pi(4k-1)}{2}\right) + \sum_{k=1}^{\infty}\frac{1 - (-1)^{4k}}{\pi (4k)}\sin\left(\frac{\pi(4k)}{2}\right)\\ &= \frac{1}{2} + \sum_{k=1}^{\infty}\frac{1 - (-1)^{4k-3}}{\pi (4k-3)} - \sum_{k=1}^{\infty}\frac{1 - (-1)^{4k-1}}{\pi (4k-1)} \\ &= \frac{1}{2} + \frac{2}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{(4k-3)} - \sum_{k=1}^{\infty}\frac{1}{(4k-1)}\right) \\ &= \frac{1}{2} + \frac{2}{\pi}\left(\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{2k-1}\right) \\ \text{shifting indices by 1, i.e. by letting $i = k-1$} \\ &= \frac{1}{2} + \frac{2}{\pi}\sum_{i=0}^{\infty}\frac{(-1)^{i}}{2i+1}. \end{align} $$ Equating the left and right hand sides then yields $$ \frac{\pi}{4} = \sum_{i=0}^{\infty}\frac{(-1)^{i}}{2i+1}, $$ precisely as required.
Note: In the above we used the fact that $$ \sin\left(\frac{\pi n}{2}\right) = \begin{cases} 1 & \text{if } n = 4k-3,\ k \in \mathbb{N} \\ 0 & \text{if } n = 4k-2,\ k \in \mathbb{N} \\ -1 & \text{if } n = 4k-1,\ k \in \mathbb{N} \\ 0 & \text{if } n = 4k,\ k \in \mathbb{N}, \end{cases} $$ for example $$ \sin\left(\frac{n\pi}{2}\right) = 1, 0, -1, 0, 1, 0, -1,... $$ for $$ n = 1,2,3,4,5,6,7... $$ respectively.