No Fibbing

It can be easily proved geometrically that $\displaystyle \sum_{n=1}^x F_{2n-1} = F_{2x}$

It can also be proved geometrically that $\displaystyle \sum_{n=1}^x ({F_{n}}^2) = F_{x}F_{x+1}$

This leaves us with the equation $\displaystyle \lim_{x\to \infty} \frac {F_{2x}}{F_{x}F_{x+1}}$

By d'Ocagne's identity, we know that $F_{2x} = F_{x}(F_{x-1} + F_{x+1})$ leaving us with the equation $\frac {F_{x}(F_{x-1} + F_{x+1})}{F_{x}F_{x+1}}$

This simplifies to $1 + \frac {F_{x-1}}{F_{x+1}}$ .

As $x \rightarrow \infty$ , The fraction becomes $1 + \frac {1}{\phi ^2}$ .

$\phi = \frac {\sqrt{5} + 1}{2}$ and therefore $1 + \frac {1}{\phi ^2}$ simplifies to $\frac {5 - \sqrt{5}}{2}$ .

Therefore $a + b + c = \boxed{12}$

A fairly non rigorous way would be to note that:

If $\lim_{x\to \infty} \frac{\sum_{n=1}^x f(n)}{\sum_{n=1}^x g(n)} = k$ , then $\lim_{x\to \infty} \frac{f(x)}{g(x)} = k$ .

$\lim_{x\to \infty} \frac{F_{2n-1}}{F_n^2} = \frac{ \frac{ \phi^{2n-1} } {\sqrt5} } { \frac{ \phi^{2n} } {5} } = \phi ^{-1} \times \sqrt5 = \frac{5 - \sqrt5}{2}$

$\begin{aligned} L & = \lim_{n \to \infty} \frac {\sum_{k=1}^n F_{2k-1}}{\sum_{k=1}^n F_k^2} & \small \color{#3D99F6} \text{Note that }\sum_{k=1}^n F_{2k-1} = F_{2n} \text{ and } \sum_{k=1}^n F_k^2 = F_n F_{n+1} \\ & = \lim_{n \to \infty} \frac {F_{2n}}{F_nF_{n+1}} & \small \color{#3D99F6} \text{Binet's formula: }F_n = \frac {\varphi^n - (-\varphi)^{-n}}{\sqrt 5} \\ & = \lim_{n \to \infty} \frac {\sqrt 5 \left(\varphi^{2n}-\varphi^{-2n}\right)}{\left(\varphi^n+\varphi^{-n}\right)\left(\varphi^{n+1}-\varphi^{-n-1}\right)} \\ & = \frac {\sqrt 5}\varphi = \frac {2\sqrt 5}{\sqrt 5+1} = \frac {5-\sqrt 5}2 \end{aligned}$

Therefore, $a+b+c = 5+5+2 = \boxed{12}$ .