More Fibonacci magic

Note that

$\begin{aligned} 1^2+1^2 &=1 \cdot 2 \\ 1^2+1^2+2^2 &=2 \cdot 3 \\ 1^2+1^2+2^2+3^2 &=3 \cdot 5 \\ 1^2+1^2+2^2+3^2 +5^2 &=5 \cdot 8 \\ \vdots \quad &= \quad \vdots \\ \displaystyle \sum_{k=1}^{n} F_k^2&= F_n \cdot F_{n+1} \end{aligned}$

(There is a nice graphical proof of this identity.)

It is commonly known that $\displaystyle\lim_{n \to \infty}\dfrac{F_{n+1}}{F_n}=\varphi$ , the golden ratio $(\varphi=\frac{1+\sqrt{5}}{2})$ .

Therefore, $\displaystyle\lim_{n \to \infty}\dfrac{\displaystyle\sum_{k=1}^{n}F_k^2}{\displaystyle\sum_{k=1}^{n-1}F_k^2}=\displaystyle\lim_{n \to \infty}\dfrac{F_n \cdot F_{n+1}}{F_{n-1} \cdot F_{n}}=\varphi^2\approx2.618033988...$

Very nice proof by Gabriel. Here is a not really nice approach though (note it is very rough approximation).

We may use the closed form of Fibonacci numbers,

$F_n = \frac{\phi^n-\psi^n}{\sqrt{5}}$

while $\phi \approx 1.6$ and $\psi \approx -0.6$ . The trick is to neglect $\psi$ as $n \rightarrow \infty$ , as $|\psi| <1$ .

$lim_{n \rightarrow \infty} \frac{\sum_{k=1}^{n}F_k^2}{\sum_{k=1}^{n-1}F_k^2}=1+\frac{F_n^2}{\sum_{k=1}^{n-1}F_k^2}$

$1+\frac{F_n^2}{\sum_{k=1}^{n-1}F_k^2}=1+\frac{\frac{\phi^{2n}}{5}}{\frac{1}{5}\sum_{k=1}^{n-1}\phi^{2k}} \approx \frac{\frac{\phi^{2n}}{1}}{\frac{\phi^{2n}-\phi^2}{\phi^2-1}} \approx \phi^2-1$

Therefore, the final solution is

$1+(\phi^2-1)=\phi^2 \approx 1.6^2$