Fibonacci infinite sum resulting in $\pi$

Question

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

2. 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 $x=\frac{i}{\phi^2}$?

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

Answer 1

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.

Answer 2

Start with $$ \begin{align} \sum_{k=0}^\infty F_{2k+1}x^{2k} &=\frac{1-x^2}{1-3x^2+x^4}\\ &=\frac{1-x^2}{(x-\phi)(x+\phi)(x+\frac1\phi)(x-\frac1\phi)}\\ &=\frac1{2\sqrt5}\frac1{x+\phi}-\frac1{2\sqrt5}\frac1{x-\phi}+\frac1{2\sqrt5}\frac1{x+\frac1\phi}-\frac1{2\sqrt5}\frac1{x-\frac1\phi}\tag{1} \end{align} $$ Integration yields $$ \begin{align} \sum_{k=0}^\infty\frac{F_{2k+1}}{2k+1}x^{2k+1} &=\frac1{2\sqrt5}\log\left(\frac{(x+\phi)(x+\frac1\phi)}{(x-\phi)(x-\frac1\phi)}\right)\\ &=\frac1{2\sqrt5}\log\left(\frac{x^2+\sqrt5x+1}{x^2-\sqrt5x+1}\right)\tag{2} \end{align} $$ Substituting $x\mapsto ix$ then multiplying by $-i\sqrt5$ gives $$ \begin{align} \sqrt5\sum_{k=0}^\infty(-1)^k\frac{F_{2k+1}}{2k+1}x^{2k+1} &=\frac1{2i}\log\left(\frac{1-x^2+i\sqrt5x}{1-x^2-i\sqrt5x}\right)\\ &=\frac1{2i}\left[\frac12\log\left(1+3x^2+x^4\right)+i\arctan\left(\frac{\sqrt5\,x}{1-x^2}\right)\right]\\ &-\frac1{2i}\left[\frac12\log\left(1+3x^2+x^4\right)-i\arctan\left(\frac{\sqrt5\,x}{1-x^2}\right)\right]\\ &=\arctan\left(\frac{\sqrt5\,x}{1-x^2}\right)\tag{3} \end{align} $$ since $\log(a+ib)=\tfrac12\log(a^2+b^2)+i\arctan\left(\tfrac ba\right)$ for $a\gt0$; that is $|x|\lt1$.

Evaluating $(3)$ at $x=\frac1{\phi^2}$ yields $$ \begin{align} \sqrt5\sum_{k=0}^\infty(-1)^k\frac{F_{2k+1}}{2k+1}\frac1{\phi^{4k+2}} &=\arctan\left(\frac{\sqrt5}{\phi^2-\frac1{\phi^2}}\right)\\ &=\arctan(1)\\[9pt] &=\frac\pi4\tag{4} \end{align} $$

Answer 3

1.Is this integration a valid thing to do? because the sum on the left has a value but the integral on the right has some constant added to it .Then how to choose the constant ?

Answer: Yes, integration is a perfectly valid step in a solution to problems like this, but not indefinite integration. The integration must be definite, otherwise the equality is between a function on the LHS:

$$ \sum_{n=0}^\infty{}F_{2n+1}\frac{x^{2n+1}}{2n+1} $$ and a set of functions on the RHS $$ \left\{c + \int\frac{1-x^2}{x^4-3x^2+1}\text{d}x,\ c\in\Bbb{R}\right\} $$ which doesn't make any sense.

So, we have to choose appropriate limits. To choose these limits, think about what you did to the LHS to go from $\sum{F_{2n+1}x^{2n}}$ to $\sum{F_{2n+1}\frac{x^{2n+1}}{2n+1}}$. We integrated the former sum from $0$ to $x$: $$ \int_0^x\sum_{n=0}^\infty{F_{2n+1}t^{2n}}\text{d}t = \sum_{n=0}^\infty{F_{2n+1}\frac{x^{2n+1}}{2n+1}} $$ but since the sum in the integrand is equal to $\frac{1-x^2}{x^4-3x^2+1}$, then we also have $$ \begin{align} \int_0^x\sum_{n=0}^\infty{F_{2n+1}t^{2n}}\text{d}t & = \int_0^x\frac{1-t^2}{t^4-3t^2+1}\text{d}t \\ & = \frac{1}{2\sqrt5}\log\left(\frac{x^2+\sqrt5x+1}{x^2-\sqrt5x+1}\right) \equiv \text{A}(x) \end{align} $$ From here, robjohn's answer explains the rest of the proof.