A Proof of Golden Section of Fibonacci Sequence

I'm glad to show you one of the feasible proof methods for the fllowing equation:

When n approaches positive infinity, then \({f_{n-1}\over f_{n}}={{\sqrt{5}-1}\over 2}\) ，which one we usually call golden section.

Fllowing, we define \(f_{i}\) be the \(i_{th}\) number of a Fibonacci sequence.

Part One: The General Formula of Fibonacci Sequence

As we all know，we defines Fibonacci sequence using \(f_{n}=f_{n-1}+f_{n-2}\)

And we also know, if we have a sequence satisfy that \(g_{i}=kg_{i-1}\), then its general formula is \(g_{i}=k^{i-1}g_{1}\)

So we transform the Fibonacci sequence into this form:

\[f_{n}-\lambda f_{n-1}=\mu (f_{n-1}-\lambda f_{n-2})\tag{0} \]

Substitute \(f_{n}=f_{n-1}+f_{n-2}\) into this equation，we will get:

\[f_{n-1}+f_{n-2}-\lambda f_{n-1}=\mu f_{n-1}-\mu\lambda f_{n-2} \]

\[(\mu +\lambda -1)f_{n-1}=(1+\mu\lambda)f_{n-2} \]

Because \(f_{n-1}\neq f_{n-2}\) , so：

\[\begin{cases}\mu +\lambda -1=0\\1+\mu\lambda=0\end{cases} \]

Solve the equation, we get the answer:

\[\begin{cases} \lambda = \frac{1+\sqrt{5}}{2}\\ \mu = \frac{1-\sqrt{5}}{2} \end{cases}\]

Or

\[\begin{cases} \lambda = \frac{1-\sqrt{5}}{2}\\ \mu = \frac{1+\sqrt{5}}{2} \end{cases}\]

Substitute it to \((0)\)

\[f_n-\frac{1+\sqrt{5}}{2}f_{n-1}=(\frac{1-\sqrt{5}}{2})^{n-2}(f_2-\frac{1+\sqrt{5}}{2}f_1)\tag{1} \]

Or

\[f_n-\frac{1-\sqrt{5}}{2}f_{n-1}=(\frac{1+\sqrt{5}}{2})^{n-2}(f_2-\frac{1-\sqrt{5}}{2}f_1)\tag{2} \]

Based on the proof just now, Both of these equations hold relative to the original equation, So we try to eliminate \(f_{n-1}\) ,then we finally get what we want:

\[f_{n}=\frac{1}{\sqrt{5}}[(\frac{\sqrt{5}+1}{2})^{n}-(\frac{1-\sqrt{5}}{2})^{n}] \]

Part Two: The Relationship of \(f_{n}f_{n-2}\) and \(f_{n-1}^{2}\)

\[f_{n}f_{n-2}-f_{n-1}^{2}=(-1)^{n-1} \]

To simplify our proof, we define that \(P=\frac{\sqrt{5}+1}{2},Q=\frac{1-\sqrt{5}}{2}\) ，then we have \(\frac{1}{\sqrt{5}}(P^{n}-Q^{n})\)

then we substitute the general formula to \(f_{n}f_{n-2}-f_{n-1}^{2}\) :

\[\frac{1}{\sqrt{5}\times\sqrt{5}}(P^{n}-Q^{n})(P^{n-2}-Q^{n-2})-\frac{1}{(\sqrt{5})^{2}}(P^{n-1}-Q^{n-1})^{2} \]

\[\frac{1}{5}[P^{2n-2}+Q^{2n-2}-P^{n}Q^{n-2}-P^{n-2}Q^{n}-(P^{2n-2}+Q^{2n-2}-2P^{n-1}Q^{n-1})] \]

\[\frac{1}{5}[-P^{n}Q^{n-2}-P^{n-2}Q^{n}+2P^{n-1}Q^{n-1}] \]

\[-\frac{1}{5}[P^{n-1}Q^{n-1}(\frac{P}{Q}+\frac{Q}{P}-2)] \]

\[-\frac{1}{5}[(PQ)^{n-1}(\frac{P}{Q}+\frac{Q}{P}-2)] \]

Remember that we defined \(P=\frac{\sqrt{5}+1}{2},Q=\frac{1-\sqrt{5}}{2}\) , so \(PQ=-1,\frac{P}{Q}=-\frac{3+\sqrt{5}}{2},\frac{Q}{P}=-\frac{3-sqrt{5}}{2},\frac{P}{Q}+\frac{Q}{P}-2=-5\) , \(-5\times -\frac{1}{5}=1\) , then we get:

\[f_{n}f_{n-2}-f_{n-1}^{2}=(-1)^{n-1} \]

Part Three: The Final Proof I

According to Part Two, \(f_{n}f_{n-2}-f_{n-1}^{2}=(-1)^{n-1}\), Transfer the term to this equation.

\[f_{n}f_{n-2}=(-1)^{n-1}+f_{n-1}^{2} \]

Then we divide both sides of the equation by \(f_{n-1}f_{n-2}\) simultaneously.

\[\frac{f_{n}}{f_{n-1}}=\frac{(-1)^{n-1}}{f_{n-1}f_{n-2}}+\frac{f_{n-1}}{f_{n-2}} \]

In our definition, \(n\) is approaching positive infinity, namely \(n\rightarrow +\infty\) , \(\frac{(-1)^{n-1}}{f_{n-1}f_{n-2}}\rightarrow 0\) , this item has a negligible impact on our answer, so we will omit it.

\[\frac{f_{n}}{f_{n-1}}=\frac{f_{n-1}}{f_{n-2}} \]

Part Four: The Final Proof II

According to Part Three, we define k that \(\frac{f_{n}}{f_{n-1}}=\frac{f_{n-1}}{f_{n-2}}=k\) , notice that \(f_{i}=f_{n-1}+f_{n-2}\)

\[\frac{f_{n-1}+f_{n-2}}{f_{n-1}}=\frac{f_{n-1}}{f_{n-2}}=k \]

\[1+\frac{f_{n-2}}{f_{n-1}}=\frac{f_{n-1}}{f_{n-2}}=k \]

\[f_{n-1}f_{n-2}+f_{n-2}^{2}=f_{n-1}^{2} \]

We define that \(f_{n-1}=kf_{n-2}\)

\[kf_{n-2}^{2}+f_{n-2}^{2}=k^{2}f_{n-2}^{2} \]

divide both sides of the equation by $$f_{n-2}^{2}$$ simultaneously.

\[k+1=k^{2} \]

Solve this equation, we finally get \(k=\frac{1+\sqrt{5}}{2}\)