Fibonacci infinite sum resulting in \pi\pi

I found the following identity. While trying to prove it, I found some things that I don’t quite understand:

\frac{\pi}{4}=\sqrt{5} \sum_{n=0}^{\infty} \frac{(-1)^n F_{2n+1}}{(2n+1) \phi^{4n+2}}

(where \phi=\frac{\sqrt{5}+1}{2}).

What I tried

I first considered the series:
F(x)=\sum_{n=0}^{\infty}F_{2n+1}x^{n}=\frac{1-x}{x^2-3x+1} (when it converges).

Then I replaced x with x^2 and tried integrating to get something like:

A(x)=\sum_{n=0}^{\infty}F_{2n+1} \frac{x^{2n+1}}{2n+1}=\int \frac{1-x^2}{x^4-3x^2+1}

This is where one question arises:

  1. Is this integration a valid thing to do? The sum on the left has a value, but the integral on the right has some constant added to it. In that case, how should I choose the constant?

Now, making x=\frac{1}{\phi^2} would make the \phi^{4n+2} term appear, but we still need to put a (-1)^n in there, so I thought of putting an i there (because i^{2n+1}=i \cdot (-1)^n) this way:

x=\frac{i}{\phi^2}, and then A(x)=i \sum_{n=0}^{\infty} \frac{(-1)^n F_{2n+1}}{(2n+1) \phi^{4n+2}}. So now I need to prove that the integral for this x is exactly \frac{i \pi}{4\sqrt{5}}, but the problem I have is that the integral has logarithms and I don’t know how to find logarithms of complex numbers like \log(5+2i). (I found on Wikipedia the Taylor series for logarithms, but I can’t see how this makes the problem simpler.)

More questions

  1. Does it make sense to plug this complex value into the power series and the integral? If so, then how one would evaluate the integral for this specific x=\frac{i}{\phi^2}?

  2. Is there any other path to prove this intriguing identity?

Answer

By the explicit formula for Fibonacci numbers it follows that:
\color{red}{S}=\color{blue}{\sqrt{5}}\sum_{n\geq 0}\frac{(-1)^n \color{blue}{F_{2n+1}}}{(2n+1)\,\color{blue}{\varphi^{4n+2}}}=\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left(\color{blue}{\frac{1}{\varphi^{2n+1}}+\frac{1}{\varphi^{6n+3}}}\right),
hence by the arctangent Taylor series and the (arc)tangent sum formulas:
\color{red}{S} = \arctan\frac{1}{\varphi}+\arctan\frac{1}{\varphi^3}=\arctan\frac{\frac{1}{\varphi}+\frac{1}{\varphi^3}}{1-\frac{1}{\varphi^4}}=\arctan 1=\color{red}{\frac{\pi}{4}}
as wanted.

Attribution
Source : Link , Question Author : Community , Answer Author : Jack D’Aurizio

Leave a Comment