Series Expansion for Pi over 8 Root 2

Theorem

 * $\displaystyle \frac \pi {8 \sqrt 2} = \sum_{n \mathop = 1}^\infty \left({-1}\right)^{n - 1} \frac {2 n - 1} {\left({4 n - 1}\right) \left({4 n - 3}\right)}$

Proof
Let $f \left({x}\right)$ be the real function defined on $\left({-\pi \,.\,.\, \pi}\right)$ as:


 * $f \left({x}\right) = \begin{cases}

\cos x & : -\pi < x < 0 \\ -\cos x & : 0 < x < \pi \end{cases}$

From Fourier Series: $\cos x$ over $\left({-\pi \,.\,.\, 0}\right)$, $-\cos x$ over $\left({0 \,.\,.\, \pi}\right)$, we have:
 * $\displaystyle f \left({x}\right) \sim -\frac 8 \pi \sum_{r \mathop = 1}^\infty \frac {r \sin 2 r x} {4 r^2 - 1}$

Setting $x = \dfrac \pi 4$, we have:

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

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

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

Hence we have: