What is the connection and the difference between the Golden Ratio and Fibonacci Sequence? The Golden Ratio, i.e.
$\varphi = \frac{1+\sqrt{5}}{2}$
and Fibonacci sequence, i.e. 
$F_n=F_{n-1}+F_{n-2}$ with the initial conditions $F_0=0$ and $F_1=1$
are clearly connected, but not perfectly so, and I'm seeking to understand this. I've been trying to read up a bit to understand the similarities and differences between these 2.
A Live Science article, for instance, says:

Around 1200, mathematician Leonardo Fibonacci discovered the unique
  properties of the Fibonacci sequence. This sequence ties directly into
  the Golden ratio because if you take any two successive Fibonacci
  numbers, their ratio is very close to the Golden ratio. As the numbers
  get higher, the ratio becomes even closer to 1.618. For example, the
  ratio of 3 to 5 is 1.666. But the ratio of 13 to 21 is 1.625. Getting
  even higher, the ratio of 144 to 233 is 1.618. These numbers are all
  successive numbers in the Fibonacci sequence.

Or, as another source put it:

The quotient of any Fibonacci number and it's predecessor approaches
  Phi, represented as ϕ (1.618), the Golden ratio.

Based on these descriptions, it sounds like the the ratio of consecutive Fibonacci numbers and the Golden Ratio converge asymptotically but are not identical (especially with the initial ratios). I would like to understand this a little bit better.
Asides from Live Science and Google, I did some preliminary research on Mathematics SE and Cross Validated. The closest question I could find was an unanswered one which primarily focused on the relationship between Arctangents and the Fibonacci sequence.
 A: If we set $\varphi = \frac{1+\sqrt{5}}{2}$ and $\bar{\varphi}=\frac{1-\sqrt{5}}{2}$, then the Binet formula for Fibonacci numbers says:
$F_n=\frac{\varphi^n-(\bar\varphi)^{ n}}{\sqrt{5}}$.
Because $\mid \bar\varphi \mid < 1$, the term $(\bar\varphi)^n$ decays fairly quickly toward $0$.  This points to why successive ratios of Fibonacci numbers get close to $\varphi$.

Note: Here is a reference for the Binet formula: http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html
A: I've seen a few proofs of this.  Here is one I like. 
We start with the definition of Fibonacci number as the sum of the previous two terms: $F(n+1) = F(n) + F(n-1)$. We can then express the ratio of successive Fibonacci terms as follows
\begin{equation}
\begin{aligned}
\frac{F(n+1)}{F(n)} & = \frac{F(n) + F(n-1)}{F(n)} \\
& = 1 + \frac{F(n-1)}{F(n)}
\end{aligned}
\label{eq:fibonacci_limit_varphi_proof1_1}
\end{equation}
We now define a quantity $x$ as
\begin{equation}
x = \lim_{n \to \infty}\ \frac{F(n+1)}{F(n)}
\end{equation}
The inverse of this quantity is 
\begin{equation}
\frac{1}{x} = \lim_{n \to \infty}\ \frac{F(n)}{F(n+1)}
\label{eq:fibonacci_limit_varphi_proof1_2}
\end{equation}
In plain english, this expression says ``take the limit as $n$ approaches infinity of a Fibonacci term and the next term.  We can simply make a change in notation and get the exact same limit:
\begin{equation}
\frac{1}{x} = \lim_{n \to \infty}\ \frac{F(n-1)}{F(n)}
\label{eq:fibonacci_limit_varphi_proof1_3}
\end{equation}
That is, if you spell out what the above limit is seeking, we'll see that
\begin{equation}
\frac{1}{x} = \lim_{n \to \infty}\ \frac{F(n)}{F(n+1)} = \lim_{n \to \infty}\ \frac{F(n-1)}{F(n)}
\label{eq:fibonacci_limit_varphi_proof1_4}
\end{equation}
Returning to the first equation and applying the limit as $n \rightarrow \infty$ to both sides, we get
\begin{equation}
\begin{aligned}
\lim_{n \to \infty}\ \frac{F(n+1)}{F(n)} & = \lim_{n \to \infty}\ \left[1 + \frac{F(n-1)}{F(n)} \right] \\
{} & = 1+ \lim_{n \to \infty}\ \frac{F(n-1)}{F(n)} \\
\end{aligned}
\end{equation}
The left-hand side is what we defined as $x$ and the limit on the right-hand side is what we discussed to be $\displaystyle \frac{1}{x}$.  So we re-write the above as
\begin{equation}
\begin{aligned}
x & = 1 + \frac{1}{x} \\
x^2 - x - 1 & = 0
\end{aligned}
\end{equation}
Solving this quadratic gives us
\begin{equation}
\begin{aligned}
\varphi & = {1 \pm \sqrt{5}  \over 2} \nonumber \\
& = 1.6180339887.... , -0.618033887...
\end{aligned}
\end{equation}
The positive root of this quadratic equation is $\varphi$.  So
\begin{equation}
x = \lim_{n \to \infty}\ \frac{F(n+1)}{F(n)} = \varphi
\end{equation}
