Obtaining a series that converges to $\frac{\pi}{2\sqrt{2}}$ using Fourier series. I'm working in some exercises about Fourier series (I'm new in that topic) but the next exercise is so hard.

Prove, finding the appropriate Fourier series, that $$\dfrac{\pi}{2\sqrt{2}}=1+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\cdots$$

First of all, I noted that the series $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{2n-1}$ is divergent by the comparison test with the divergent series $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{2n}$. Then, how can the series converges to $\dfrac{\pi}{2\sqrt{2}}$? Or am I missing something?
Then, I tried to follow the steps in this answer but I can't because I don't have any function or an option to use. How can I do? Any hint? I really appreciate any help you can give me.
 A: 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.
