Series Expansion for Pi over 8 Root 2

Theorem

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

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


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

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

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

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: