Series Expansion for Pi over Root 2/Proof 1

Proof
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:


 * $\map f x = \begin{cases}

\sin \dfrac x 2 & : 0 \le x < \pi \\ -\sin \dfrac x 2 & : \pi < x \le 2 \pi \end{cases}$

From Fourier Series: $\sin \dfrac x 2$ over $\closedint 0 \pi$, $-\sin \dfrac x 2$ over $\closedint 0 {2 \pi}$, we have:


 * $\map f x \sim \displaystyle \frac 8 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {n \sin n x} {4 n^2 - 1}$

Setting $x = \dfrac {\pi} 2$, we have:

When $n$ is even, $\dfrac {n \pi} 2$ is an integer multiple of $\pi$.

Hence, in this case, from Sine of Multiple of Pi:
 * $\sin \dfrac {n \pi} 2 = 0$

When $n$ is odd it can be expressed as $n = 2 r - 1$ for $r \ge 1$.

Hence we have: