Obtaining a series that converges to $\frac{\pi}{2\sqrt{2}}$ using Fourier series.

First we have to get the signs right in the series or else bad things happen, as described in the comments. Properly, only terms with the denominator $\in\{1,3\}\bmod 8$ are positive. Where the denominator is $\in\{5,7\}\bmod 8$ the terms are negative. The corrected series is now in the question.

As you probably know, the more familiar series expansion

$\dfrac{\pi}{4}=1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+...$

may be proved by rendering the odd periodic extension of

$f(x)=1,0\le x<\pi$

and evaluating the resulting Fourier series at $x=\pi/2$. Can we find a similar odd periodic extension that contains $\sqrt2$ and thereby get a result involving $\pi/\sqrt2$?

Knowing that $\sin(\pi/4)=(\sqrt2)/2$, let us try an odd periodic extension of

$f(x)=\sin(x/2),0\le x<\pi$

with the intent of again putting $x=\pi/2$ into the result.

Our Fourier series for the odd periodic extension is then

$F(x)=\Sigma_{n=0}^\infty s_n\sin nx$

$s_n=\dfrac{1}{\pi}\int_0^\pi \sin(x/2)\sin(nx)dx$

We apply the appropriate trigonometric sum-product relation:

$s_n=\dfrac{1}{\pi}\int_0^\pi (\cos((n-\frac{1}{2})x)-\cos((n+\frac{1}{2})x))dx$

$=\dfrac{2}{(2n-1)\pi}\sin((n-\frac{1}{2})\pi)-\dfrac{2}{(2n+1)\pi}\sin((n+\frac{1}{2})\pi)$

Render $\sin((n-\frac{1}{2})\pi)=-1$ for $n$ even but $+1$ for $n$ odd, and the reverse for $\sin((n+\frac{1}{2})\pi)$:

$s_n=\dfrac{2}{\pi}(-1)^{n-1}\left(\dfrac{1}{2n-1}+\dfrac{1}{2n+1}\right)$

So

$F(x)=\dfrac{2}{\pi}\Sigma_{n=0}^\infty (-1)^{n-1}\left(\dfrac{1}{2n-1}+\dfrac{1}{2n+1}\right)\sin nx$

And plug in $x=\pi/2$ to get the sum:

$\dfrac{\sqrt2}{2}=\dfrac{2}{\pi}\left(\left(1+\dfrac{1}{3}\right)(1)-\left(\dfrac{1}{3}+\dfrac{1}{5}\right)(0)+\left(\dfrac{1}{5}+\dfrac{1}{7}\right)(-1)-\left(\dfrac{1}{7}+\dfrac{1}{9}\right)(0)+...\right)$

$\dfrac{\sqrt2}{2}=\dfrac{2}{\pi}\left(\left(1+\dfrac{1}{3}\right)-\left(\dfrac{1}{5}+\dfrac{1}{7}\right)+\left(\dfrac{1}{9}+\dfrac{1}{11}\right)-...\right)$

from which the claimed result (with proper signs) follows via a simple algebraic rearrangement.