Series Expansion for Pi over Root 2

Theorem

 * $\displaystyle \frac \pi {\sqrt 2} = \sum_{r \mathop = 1}^\infty \left({-1}\right)^{r - 1} \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} }$

Proof
Let $f \left({x}\right)$ be the real function defined on $\left({0 \,.\,.\, 2 \pi}\right)$ as:


 * $f \left({x}\right) = \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 $\left[{0 \,.\,.\, \pi}\right]$, $-\sin \dfrac x 2$ over $\left[{\pi \,.\,.\, 2 \pi}\right]$, we have:


 * $f \left({x}\right) \sim \displaystyle \frac 8 \pi \sum_{n \mathop = 1}^\infty \left({-1}\right)^{n - 1} \frac {n \sin n x} {4 n^2 - 1}$

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: