Prove the Identity $\pi=2- \sum_{1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}} $

Question

By considering the fact that $f(\pi/2)=1$, prove the identity

$\pi=2- \sum_{1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}} $

This question was is a subsection in a chapter on Fourier series, can I use my understanding of Fourier series to prove the identity, or is there an easier way of making the proof?

EDIT Another part of this question asked to find the Fourier series of the periodic function $f(x)$ defined by $f(x)= |sin(x)|$, this is the function they are probably referring to in the proof.

Presumably you need to take whatever $f$ is, find its series expansion, plug in $\pi/2$, and then maybe do some algebra. — Ian, Aug 20 '15 at 22:02
I suppose using complex variables $$h(z) = \frac{\pi}{\sin(\pi z)}\frac{1}{z^2-1/4}$$ should work well here. This is not the same as the function referenced by the OP. — Marko Riedel, Aug 20 '15 at 22:09
The function which is to be used is $f(x)=|sin(x)|$, when I calculate its Fourier series expansion, what would be the next step in equating it to a summation involving an $m$ constant? @Ian — mnmakrets, Aug 20 '15 at 22:10
How would I use this function to prove the identity? @MarkoRiedel — mnmakrets, Aug 20 '15 at 22:11
So you have the Fourier series, and probably know that it converges pointwise to $f$, Thus plug $x = \frac{\pi}{2}$ into $$f(x) = \frac{a_0}{2} + \sum_{n = 1}^\infty a_n \cos (nx).$$ — Daniel Fischer, Aug 20 '15 at 22:14
The comment above refers to this technique (MSE link) which admittedly does not use Fourier series. (Compute all residues at the poles of $h(z)$ and show that the integral along the contour vanishes in the limit.) — Marko Riedel, Aug 20 '15 at 22:25
You can prove this only using the Gregory-Leibniz series for $\frac\pi4$. (I'm referring to $\frac\pi4=1-\frac13+\frac15-\dotsb$.) — Akiva Weinberger, Aug 20 '15 at 23:58
@columbus8myhw: very true. It is enough to rearrange a little the RHS of the first line of my proof. — Jack D'Aurizio, Aug 21 '15 at 00:03

score 12 · Answer 1 · answered Aug 20 '15 at 22:21

12

Fourier series are not really needed to prove the identity: $$\begin{align*}\sum_{m\geq 1}\frac{(-1)^m}{4m^2-1}&=\frac{1}{2}\sum_{m\geq 1}(-1)^m\left(\frac{1}{2m-1}-\frac{1}{2m+1}\right)\\&=\frac{1}{2}\sum_{m\geq 1}(-1)^m\int_{0}^{1}\left(x^{2m-2}-x^{2m}\right)\,dx\\&=\frac{1}{2}\int_{0}^{1}\frac{x^2-1}{x^2+1}\,dx\\&=\frac{1}{2}\left(1-2\arctan 1\right)\\&=\color{red}{\frac{2-\pi}{4}}.\end{align*}$$

answered Aug 20 '15 at 22:21

Jack D'Aurizio

353,855

Very nice. Much simpler than mine. – marty cohen Aug 20 '15 at 22:39
1

I just posted a solution for a tedious integral and received a comment that a wise young man once said, and I'm paraphrasing here, that there is more than 1 way to skin a series. – Mark Viola Aug 20 '15 at 22:41
Can anyone explain how to get from the second line to the third? – Teman Aug 20 '15 at 23:39
1

@hans__ You mean, how he got rid of the $\sum$? Switch $\sum$ and $\int$, and use the geometric series. – Akiva Weinberger Aug 20 '15 at 23:56
@columbus8myhw how is the switch possible? And we have $0\leq x\leq 1$ instead of $0\leq x <1$, no? I am mindful of the gap in my knowledge of analysis. – Teman Aug 20 '15 at 23:58
@hans__: a single point does not matter for integration purposes. $x^{2m-2}-x^{2m}$ is a non-negative and bounded function on $(0,1)$ having integral $O\left(\frac{1}{m^2}\right)$, so we are more than allowed to switch the series and the integral (we have absolute convergence). – Jack D'Aurizio Aug 21 '15 at 00:01
You could have separated the series, made a substitution, and recombined, rather than messing with integrals, I believe. It would lead you straight to the Gregory-Leibniz formula. – Akiva Weinberger Aug 21 '15 at 00:02
@columbus8myhw: I do not think my approach is "messy" but you are right, that is an even shorter way. – Jack D'Aurizio Aug 21 '15 at 00:04
"Messing," not "messy." I didn't mean to imply anything. – Akiva Weinberger Aug 21 '15 at 00:06

score 3 · Answer 2 · answered Aug 20 '15 at 22:35

I'll play around and see what happens.

We want $\pi=2- \sum_{1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}} $

$\begin{array}\\ S &=\sum_{m=1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}}\\ &=\sum_{m=1}^{∞} \frac{(-1)^m}{m^2}\frac1{1-\frac{1}{4m^2}}\\ &=\sum_{m=1}^{∞} \frac{(-1)^m}{m^2}\sum_{j=0}^{\infty}\frac{1}{(4m^{2})^j}\\ &=4\sum_{m=1}^{\infty}(-1)^m\sum_{j=0}^{\infty}\frac{1}{(4m^{2})^{j+1}}\\ &=4\sum_{m=1}^{\infty}(-1)^m\sum_{j=1}^{\infty}\frac{1}{(4m^{2})^{j}}\\ &=4\sum_{j=1}^{\infty}\sum_{m=1}^{\infty}(-1)^m\frac{1}{(4m^{2})^{j}}\\ &=4\sum_{j=1}^{\infty}\frac1{4^j}\sum_{m=1}^{\infty}(-1)^m\frac{1}{(m^{2})^{j}}\\ &=4\sum_{j=1}^{\infty}\frac1{4^j}\sum_{m=1}^{\infty}(-1)^m\frac{1}{m^{2j}}\\ &=-4\sum_{j=1}^{\infty}\frac1{4^j}\zeta(2j)(1-2^{1-2j})\\ &=-4\sum_{j=1}^{\infty}\frac1{4^j}\zeta(2j)+8\sum_{j=1}^{\infty}\frac1{4^{2j}}\zeta(2j))\\ \end{array} $

According to https://en.wikipedia.org/wiki/Riemann_zeta_function, $\sum_{j=1}^{\infty}\frac{\zeta(2j)-1}{4^j} =\frac16 $ and $\sum_{j=1}^{\infty}\frac{\zeta(2j)-1}{16^j} =\frac{13}{30}-\frac{\pi}{8} $. Therefore

$\begin{array}\\ \sum_{j=1}^{\infty}\frac{\zeta(2j)}{4^j} &=\frac16+\sum_{j=1}^{\infty}\frac{1}{4^j}\\ &=\frac16+\frac{1/4}{1-1/4}\\ &=\frac16+\frac13\\ &=\frac12\\ \end{array} $

and

$\begin{array}\\ \sum_{j=1}^{\infty}\frac{\zeta(2j)}{16^j} &=\frac{13}{30}-\frac{\pi}{8}+\sum_{j=1}^{\infty}\frac{1}{16^j}\\ &=\frac{13}{30}-\frac{\pi}{8}+\frac{1/16}{1-1/16}\\ &=\frac{13}{30}-\frac{\pi}{8}+\frac{1}{15}\\ &=\frac12-\frac{\pi}{8}\\ \end{array} $.

Putting this together,

$\begin{array}\\ S &=-4\sum_{j=1}^{\infty}\frac1{4^j}\zeta(2j)+8\sum_{j=1}^{\infty}\frac1{4^{2j}}\zeta(2j)\\ &=-4(\frac12)+8(\frac12-\frac{\pi}{8})\\ &=-2+4-\pi\\ &=2-\pi\\ \end{array} $.

Whew!

Even though someone posted a simpler solution, +1 for typing all of that out — Brenton, Aug 20 '15 at 22:41
@user109899: I do not think so. With marty's conversion it is enough to exploit $$\forall x\in(-1,1),\qquad \sum_{n\geq 1}\zeta(2n),x^{2n}=\frac{1-\pi x\cot(\pi x)}{2}.$$ — Jack D'Aurizio, Aug 21 '15 at 00:48

Prove the Identity $\pi=2- \sum_{1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}} $

2 Answers2