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:

- 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

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}?

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*