Limit of a function not using Stirling's Approximation

Question

I want to compute the following limit: $$\lim_{n\to\infty} \frac{\left(\frac{e}{F_{n+1}}\right)^{F_{n+1}} F_{n+1}!}{\left(\frac{e}{F_n}\right)^{F_n} F_n!},$$

where $F_n$ is the $n$th Fibonacci number. The limit is easily computed by using Stirling's approximation $n! \simeq \sqrt{2\pi n} \left(\frac{n}{e}\right)^n$:

$$\lim_{n\to\infty} \frac{\left(\frac{e}{F_{n+1}}\right)^{F_{n+1}} F_{n+1}!}{\left(\frac{e}{F_n}\right)^{F_n} F_n!} = \lim_{n\to\infty} \frac{\left(\frac{e}{F_{n+1}}\right)^{F_{n+1}} \sqrt{2\pi F_{n+1}} \left(\frac{F_{n+1}}{e}\right)^{F_{n+1}}}{\left(\frac{e}{F_n}\right)^{F_n} \sqrt{2\pi F_n} \left(\frac{F_n}{e}\right)^{F_n}}\\=\lim_{n\to\infty}\sqrt{\frac{F_{n+1}}{F_n}}\\=\sqrt{\frac{1+\sqrt{5}}{2}}.$$

Is it possible to show this without using Stirling's approximation?

Answer 1

$$(\frac e{F_{n+1}})^{F_{n+1}}F_{n+1}!\over (\frac e{F_n})^{F_n}F_n!$$

$$=\frac {\frac {e^{F_{n+1}-F_n}}{F_{n+1}^{F_{n+1}}}\prod_{i=F_n+1}^{F_{n+1}}i}{1\over F_n^{F_n}}$$

$$=\frac {F_n^{F_n}}{F_{n+1}^{F_{n+1}}}e^{F_{n+1}-F_n}\prod_{i=F_n+1}^{F_{n+1}}i$$

$$=\left(\frac {F_n}{F_{n+1}}\right)^{F_n}\prod_{i=F_n+1}^{F_{n+1}}\frac {ei}{F_{n+1}}\tag 1$$

Taking what we know: $F_{n+1}=F_n+F_{n-1},\ (1)$ becomes

$$\left(\frac {F_n}{F_n+F_{n-1}}\right)^{F_n}\cdot\prod_{i=F_n+1}^{F_n+F_{n-1}}\frac {ei}{F_n+F_{n-1}}$$

$$=e\left(\frac {F_n}{F_n+F_{n-1}}\right)^{F_n}\cdot\prod_{i=1}^{F_{n-1}-1}\frac {e(i+F_n)}{F_n+F_{n-1}}$$

I don't see anything that makes it much easier from here, except to note that

$$\left(\frac {F_n}{F_{n+1}}\right)^{F_n}\to \left(\frac {\sqrt 5-1}{2}\right)^{F_n}\to \frac 1{\left(\frac {1+\sqrt 5}2\right)F_n+F_{n-1}}$$

At the end of all this, that product is still waiting, and probably just needs Stirling's Approximation to complete the evaluation.