# Verification of proof involving logarithms and limits

Question: Consider $E_n\in(0,1)$ and $E_n\to0,\text{as }n\to\infty$

Assuming that $E_{n+1} = E_n E_{n-1}$, show that, for a real $C$ independent of $n$, $$E_{n+1} \leq C{E_n}^\phi$$ where $\phi=\frac{1+\sqrt{5}}{2}$, the Golden Ratio

My attempt so far:

We can take logarithms of both side to see that

\begin{align*}\log(E_{n+1}) &= \log(E_n E_{n-1})\\ &=\log(E_n)+\log(E_{n-1})\end{align*}

We can see that this fulfils the conditions of the Fibonacci Sequence, $F_{n+1} = F_{n}+F_{n-1}$

Therefore, we can assume that $$\frac{\log(E_{n+1})}{\log(E_n)}\to \phi,\quad\text{as }n\to\infty$$

We can then say that \begin{align*}\lim_{n\to\infty}\left|\frac{\log(E_{n+1})}{\log(E_n)}-\phi\right|&=0\\ \lim_{n\to\infty}\left|\log(E_{n+1})-\phi\log(E_n)\right|&=0\\ \lim_{n\to\infty}\left|\log(E_{n+1}) -\log({E_n}^\phi)\right|&=0\\ \lim_{n\to\infty}\left|E_{n+1} - {E_n}^\phi\right|&=1\end{align*}

I'm not entirely sure that the last line of this is correct - I have attempted to get rid of the logarithms as I would have done in an equation with no limits or absolute values in. Could someone let me know if I have done anything wrong please. I am also struggling to continue from this point so any guidance would be much appreciated too.

• I don't think that you can replace $\log(E_{n+1})-\log(E_n^\phi)$ like that. – Simply Beautiful Art Dec 5 '17 at 14:35
• I also don't think that you can multiply both sides by $\log(E_n)$ near the bottom, since $\lim_{n\to\infty}\log(E_n)=\log(0^+)=-\infty$. – Simply Beautiful Art Dec 5 '17 at 14:37
• @SimplyBeautifulArt Thanks for your input, how would you propose I tackle this question then? – lioness99a Dec 5 '17 at 14:39

Hint : The following functional equation : $$f(n+1)=f(n)f(n-1)$$ Have to solution : $$f(n)=e^{c_1F_n+c_2L_n}$$ Where $F_n$ are the Fibonnaci number , $L_n$ the Lucas number and $c_1$,$c_2$ are arbitrary parameters.