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

Question

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.

Answer 1

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.