Asymptotic value of Fibonacci numbers 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 $\lim_{n\to\infty}u_n$.

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