Is the convergence linear, superlinear, quadratic?

Question

The Fibonacci sequence is generated by the formulas \begin{cases} r_0=1 & r_1=1\\ r_{n+1}=r_n+r_{n-1} \end{cases} The sequence therefore starts out $1, 1, 2, 3, 5, 8, 13, 21, 34, \dots$. Prove that the sequence $[2r_n/r_{n-1}]$ converges to $1+\sqrt{5}$. Is the convergence linear, superlinear, quadratic?

Proof: Suppose we are given the following recurrence relation $$r_{n+1}=r_n+r_{n-1},$$ where $r_0=1$ and $r_1=1$. We want to show that the sequence $[2r_n/r_{n-1}]$ converges to $1+\sqrt{5}$. i.e. We want to show that $$\lim_{n\to \infty} \dfrac{2r_n}{r_{n-1}}=1+\sqrt{5}.$$ Before, we prove this limit is true, we need to find the solution to this recurrence relation with the given initial conditions.

The corresponding characteristic polynomial to the recurrence relation is $$p(\lambda)=\lambda^2-\lambda-1=0.$$ Hence the roots of the characteristic polynomial are $\lambda=\dfrac{1\pm\sqrt{5}}{2}$. Hence the general solution for the relation is $$r_n=A\left(\dfrac{1+\sqrt{5}}{2}\right)^n+B\left(\dfrac{1-\sqrt{5}}{2}\right)^n.$$

Next, we want to find the values of $A$ and $B$ with the given initial conditions. With the starting values $r_0=1$ and $r_1=1$, we get the following system of equations: \begin{cases} 1=r_0=A+B \\ 1=r_1=A\left(\dfrac{1+\sqrt{5}}{2}\right)+B\left(\dfrac{1-\sqrt{5}}{2}\right) \end{cases}

Multiply the first equation by $\dfrac{1+\sqrt{5}}{2}$. \begin{cases} \dfrac{1+\sqrt{5}}{2}=r_0=A\left(\dfrac{1+\sqrt{5}}{2}\right)+B\left(\dfrac{1+\sqrt{5}}{2}\right) \\ 1=r_1=A\left(\dfrac{1+\sqrt{5}}{2}\right)+B\left(\dfrac{1-\sqrt{5}}{2}\right) \end{cases}

Next, subtract equation (1) from equation (2), which results with $$\dfrac{\sqrt{5}=1}{2}=B\sqrt{5} \iff B=\dfrac{\sqrt{5}-1}{2\sqrt{5}}.$$ Furthermore $$A=1-\dfrac{\sqrt{5}-1}{2\sqrt{5}}=1-\dfrac{\sqrt{5}}{2\sqrt{5}}+\dfrac{1}{2\sqrt{5}}=1-\dfrac{1}{2}+\dfrac{1}{2\sqrt{5}}=\dfrac{1}{2}+\dfrac{1}{2\sqrt{5}}=\dfrac{\sqrt{5}+1}{2\sqrt{5}}.$$ Hence, the general solution for the relation is \begin{equation*} \begin{aligned} r_n & =\left(\dfrac{\sqrt{5}+1}{2\sqrt{5}}\right)\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\left(\dfrac{\sqrt{5}-1}{2\sqrt{5}}\right)\left(\dfrac{1-\sqrt{5}}{2}\right)^n \\ & =\dfrac{1}{\sqrt{5}}\left[\dfrac{(1+\sqrt{5})^{n+1}-(1-\sqrt{5})^{n+1}}{2^{n+1}}\right]. \end{aligned} \end{equation*}

Now we can show that the sequence $[2r_n/r_{n-1}]$ converges to $1+\sqrt{5}$. \begin{equation*} \begin{aligned} \lim_{n\to \infty} \dfrac{2r_n}{r_{n-1}} & = \lim_{n\to \infty} \dfrac{\dfrac{2}{\sqrt{5}}\left[\dfrac{(1+\sqrt{5})^{n+1}-(1-\sqrt{5})^{n+1}}{2^{n+1}}\right]}{\dfrac{1}{\sqrt{5}}\left[\dfrac{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}}{2^{n}}\right]} \\ & = \lim_{n\to \infty} \dfrac{(1+\sqrt{5})^{n+1}-(1-\sqrt{5})^{n+1}}{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}} \\ & = \lim_{n\to \infty} \dfrac{\dfrac{(1+\sqrt{5})^{n+1}-(1-\sqrt{5})^{n+1}}{(1+\sqrt{5})^n}}{\dfrac{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}}{(1+\sqrt{5})^n}} \\ & = \lim_{n\to \infty} \dfrac{(1+\sqrt{5})-(1-\sqrt{5})\cdot \left(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\right)^{n}}{1-\left(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\right)^{n}} \\ & = \dfrac{(1+\sqrt{5})-(1-\sqrt{5})\cdot 0}{1-0} \\ & = 1+\sqrt{5}. \end{aligned} \end{equation*}

Note: Since $0<\dfrac{1-\sqrt{5}}{1+\sqrt{5}}<1$, then $$\lim_{n\to\infty} \left(\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\right)^{n}=0.$$

To finish the question I need to state whether the convergence is linear, superlinear, or quadratic. I know that

• If the limit equals 0, then the convergence is superlinear.
• If the limit equals 1, then the convergence is linear.
• If the limit equals 2, then the convergence is quadratic.

However, my limit is equal to $1+\sqrt{5}$. Am I missing something?

Answer 1

Trying to match your notation: Let $r_n$ represent the $n$'th Fibonacci number, let $q_n=\frac{2r_n}{r_{n-1}}$, and let $L=1+\sqrt{5}$

In the first part of the problem, you should have shown $\lim\limits_{n\to\infty}q_n=L$

We now investigate what happens to the limit:

$$(\dagger):~~~~~~~~~\lim\limits_{n\to\infty}\frac{|q_{n+1}-L|}{|q_{n}-L|}$$

$q_{n+1}$ will tend to be closer to $L$ than $q_{n}$ implying that $|q_{n+1}-L|<|q_{n}-L|$ implying that the limit $(\dagger)$ above will if it exists be somewhere in the range $[0,1]$. If the limit is $1$ we say that $q_n$ converges to $L$ "sublinearly", if the limit is between $0$ and $1$ we say that $q_n$ converges to $L$ "linearly", and if the limit is $0$ we say that $q_n$ converges to $L$ "superlinearly." In the case that it is sublinear or superlinear, we can futher classify just how quickly or slowly that is by raising the denominator to various powers and investigating those limits.

Now... again, we are tasked with investigating the limit $(\dagger)$. Remembering that $q_n=\frac{2r_n}{r_{n-1}}$ and that $r_n=\frac{\varphi^n-\psi^n}{\sqrt{5}}$ where $\varphi=\frac{1+\sqrt{5}}{2}$ and $\psi=\frac{1-\sqrt{5}}{2}$. Notice that $\frac{L}{2}=\varphi$ and that $\varphi\cdot \psi=-1$.

We have then $q_n=2\cdot \frac{\varphi^n-\psi^n}{\varphi^{n-1}-\psi^{n-1}}$. We have now $q_{n+1}-L=2\cdot \frac{\varphi^{n+1}-\psi^{n+1}-\varphi(\varphi^n-\psi^n)}{\varphi^n-\psi^n}$

$=2\cdot \frac{\varphi^{n+1}-\psi^{n+1}-\varphi^{n+1}-\psi^{n-1}}{\varphi^n-\psi^n}=2\cdot \frac{-\psi^{n+1}-\psi^{n-1}}{\varphi^n-\psi^n}$

Similarly we calculate $q_{n}-L$ to be $2\cdot \frac{-\psi^n-\psi^{n-2}}{\varphi^{n-1}-\psi^{n-1}}$

As a sanity check here, notice that $\lim\limits_{n\to\infty}|q_n-L|=0$ as the numerator approaches zero while the denominator grows large. This confirms, or rather rederives, what we should have learned in the earlier part of the question.

So this leads us to our limit $(\dagger)$ being:

$\lim\limits_{n\to\infty}\frac{|q_{n+1}-L|}{|q_n-L|}=\lim\limits_{n\to\infty}\frac{|2\cdot \frac{-\psi^{n+1}-\psi^{n-1}}{\varphi^n-\psi^n}|}{|2\cdot \frac{-\psi^n-\psi^{n-2}}{\varphi^{n-1}-\psi^{n-1}}|}=$

Which after a bit more algebra and a little bit of handwaving results in

$\frac{\frac{|\psi^2+\psi|}{|\psi+1|}}{\varphi}=\frac{|\psi|}{\varphi}=\frac{\sqrt{5}-1}{1+\sqrt{5}}\approx 0.382$

The handwaving comes into play when we try to move the different parts of the fraction around to get like groups together. The group of the form $\frac{\varphi^n-\psi^n}{\varphi^{n-1}-\psi^{n-1}}$ we notice that as $\lim\limits_{n\to\infty}\psi^n=0$ the only terms that truly survive are the $\varphi'$s. This can be made more formal, but this suffices for now.

I.e. twice the ratio of consecutive fibonacci terms converges to $L$ linearly.