Asymptotic value of Fibonacci numbers

Question

It is well known that $F_n\sim\frac{\phi^n}{\sqrt{5}}$, where $\phi=\frac{1+\sqrt{5}}{2}$. Does someone know a better estimate? With proof please.

I'm trying to calculate the following limit:

Let $u_1=1, u_2=C, u_3=C$ and

$$ u_n=\frac{C^{F_{n-1}}}{2^{F_{n-3}}3^{F_{n-4}}\cdots(n-2)^{F_1}} $$ for $n\geq 4$, where

$$ C=\exp\left(\sum_{k=1}^{\infty}\frac{\log k}{\phi^k}\right). $$

Find $\lim_{n\to\infty}u_n$.

Answer 1

If we let

$$ a_n = 2^{\large F_{n-3}} 3^{\large F_{n-4}} \cdots (n-2)^{\large F_{1}} = \prod_{k=2}^{n-2} k^{\large F_{n-1-k}} $$

then

$$ \log a_n = \sum_{k=2}^{n-2} F_{n-1-k} \log k. \tag{1} $$

Using Binet's formula

$$ F_m = \frac{\varphi^m - (-\varphi)^{-m}}{\sqrt{5}}, $$

$(1)$ becomes

$$ \begin{align} \sqrt{5} \log a_n &= \varphi^{n-1} \sum_{k=2}^{n-2} \varphi^{-k} \log k - (-\varphi)^{1-n} \sum_{k=2}^{n-2} (-\varphi)^{k} \log k \\ &= b_n + c_n, \end{align} $$

where

$$ b_n = \varphi^{n-1} \sum_{k=2}^{n-2} \varphi^{-k} \log k \qquad \text{and} \qquad c_n = -(-\varphi)^{1-n} \sum_{k=2}^{n-2} (-\varphi)^{k} \log k. \tag{2} $$

The sum in $b_n$ converges as $n \to \infty$, so we'll rewrite it as

$$ b_n = \varphi^{n-1} \sum_{k=2}^{\infty} \varphi^{-k} \log k - \varphi^{n-1} \sum_{k=n-1}^{\infty} \varphi^{-k}\log k. \tag{3} $$

Lemma 1: $$ \sum_{k=n-1}^{\infty} \varphi^{-k}\log k = \frac{\varphi^{2-n}\log n}{\varphi - 1} + O\left(\frac{\varphi^{-n}}{n}\right). \tag{4} $$ Idea of the proof: Split the sum at $k=\frac{3}{2}n$. For the sum over $k \geq \frac{3}{2} n$ use the rough estimate $\log k < k$, and for the sum over $n-1 \leq k < \frac{3}{2} n$ expand $\log k$ in series about the point $k = n-1$.

Lemma 2: $$ \sum_{k=2}^{n-2} (-\varphi)^{k}\log k = \frac{-(-\varphi)^{n-1}\log n}{\varphi + 1} + O\left(\frac{\varphi^{n}}{n}\right). \tag{5} $$ Idea of the proof: Split the sum at $k=\frac{1}{2}n$ and proceed as in Lemma 1.

Combining $(2)$, $(3)$, $(4)$, and $(5)$ we obtain

$$ \sqrt{5} \log a_n = \varphi^{n-1} \sum_{k=2}^{\infty} \varphi^{-k} \log k + \left(\frac{1}{\varphi + 1} - \frac{\varphi}{\varphi - 1}\right) \log n + O\left(\frac{1}{n}\right). $$

Since $\frac{1}{\varphi + 1} - \frac{\varphi}{\varphi - 1} = -\sqrt{5}$ we can rewrite this as

$$ \log a_n = \frac{\varphi^{n-1}}{\sqrt{5}} \sum_{k=2}^{\infty} \varphi^{-k} \log k - \log n + O\left(\frac{1}{n}\right). $$

Exponentiating this yields

$$ \begin{align} a_n &= \frac{1}{n} \exp\left(\frac{\varphi^{n-1}}{\sqrt{5}} \sum_{k=2}^{\infty} \varphi^{-k} \log k\right)\left[1 + O\left(\frac{1}{n}\right)\right] \\ &= \frac{1}{n} \exp\left(F_{n-1} \sum_{k=2}^{\infty} \varphi^{-k} \log k\right)\left[1 + O\left(\frac{1}{n}\right)\right] \\ &= \frac{C^{\large F_{n-1}}}{n} \left[1 + O\left(\frac{1}{n}\right)\right], \end{align} $$

where in the second line we used the estimate $F_n = \frac{\varphi^n}{\sqrt{5}} + O(\varphi^{-n})$. Finally we have

$$ u_n = \frac{C^{\large F_{n-1}}}{a_n} = n \left[1 + O\left(\frac{1}{n}\right)\right] = n + O(1). $$