Okay, I was messing around on Excel with some coefficients and I stumbled onto this. Not sure if it converges but it gets pretty damn close around the 1024th term mark. Was wondering if somebody could either prove this to me or at least tell me whether the function is correct. Just thought it was interesting.
ϕ/(ln(2^(n-1)))/f(n))/n) - (ln(2^(n-1)))/f(n))/n) = e^2
Where ϕ is the golden ratio and f(n) is the 'n'th Fibonacci number
It may be painfully simple to the math geniuses out there but I'd love to see why.
From your comment, I suppose this is what you have asked:
\begin{align*} \lim_{n\to\infty} \frac{\phi \, n}{\ln\left(\frac{2^{n - 1}}{\mbox{f}(n)}\right)} - \frac{\ln\left(2^{n - 1}\right)}{\mbox{f}(n)\, n} \tag 1 \end{align*}
Using Binet's formula, for large $n$,
\begin{align*} f(n)\sim \frac{\phi^n}{\sqrt{5}} \end{align*}
Using that in $(1)$,
\begin{align*} &\lim_{n\to\infty} \left(\frac{\phi\, n}{(n-1) \ln{2} - n\, \ln \phi + \ln\sqrt 5} - \frac{(n-1)\, \ln 2}{\displaystyle \frac{\phi^n}{\sqrt{5}}\, n}\right)\\ &= \frac{\phi}{\ln{2}-\ln{\phi}}\approx 7.63456377974314 \end{align*}