Fibonacci infinite sum resulting in $\pi$

Solution 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.

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

Solution 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.