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I am trying to solve this problem imagine that

\begin{array}{lr} f(n) = 2^{\dfrac{1}{\sqrt{5}}\left[\left(\dfrac{1 + \sqrt{5}}{2}\right)^{n} - \left(\dfrac{1 - \sqrt{5}}{2}\right)^{n}\right]} \\ \\ g(n) = 2^{\left(\dfrac{1 + \sqrt{5}}{2}\right)^{n+1}} \end{array}

and I want to find

\begin{equation*} \lim_{n\to \infty} \dfrac{f(n)}{g(n)} \end{equation*}

now in the middle of proof I stock on this limit, how can I solve it? ($\varphi$ is the golden number)

\begin{equation*} \lim_{n\to \infty} -\frac{3+\sqrt{5}}{2} \varphi^n - \frac{1}{\sqrt{5}}(-\varphi)^{-n} \end{equation*}

Karo
  • 469

1 Answers1

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For simplicity, I will let $\varphi$ denote the golden ratio, $\overline \varphi$ its conjugate, and $F(n)$ the $n^{th}$ Fibonacci number. That is,

$$\varphi = \frac{1 + \sqrt 5}{2} \;\;\;\;\; \overline \varphi = \frac{1 - \sqrt 5}{2} \;\;\;\;\; F(n) = \frac{\varphi^n - \overline \varphi^n}{\sqrt 5}$$

Then we have

$$f(n) = 2^{F(n)} \;\;\;\;\; g(n) = 2^{\varphi^{n+1}}$$

and seek the limit as $n \to \infty$ of $f(n)/g(n)$.

We note that, owing to properties of exponents,

$$\frac{f(n)}{g(n)} = 2^{F(n) - \varphi^{n+1}}$$

Then, by continuity,

$$L := \lim_{n \to \infty} \frac{f(n)}{g(n)} = 2^{\lim \limits_{n \to \infty} F(n) - \varphi^{n+1}}$$

Thus, it will be best for us to find the limit in the exponent. Going forward we will focus entirely on that exponent as well, and use the Binet formula for $F(n)$ as established above.

We see that

$$F(n) - \varphi^{n+1} = \frac{\varphi^n}{\sqrt 5} - \frac{\overline \varphi^n}{\sqrt 5} - \varphi^{n+1}$$

Swap the last two terms and then factor out $\varphi^n$, simplifying its coefficient, and we get the below.

$$F(n) - \varphi^{n+1} = \varphi^n \left( -\frac 1 2 - \frac{3}{2 \sqrt 5} \right) - \frac{\overline \varphi^n}{\sqrt 5}$$

From here, it's pretty easy to conclude our answer already. Why? We note that $\overline \varphi \approx -0.618 \in (-1,1)$. Thus, as $n \to \infty$, that term will go to zero. So, effectively,

$$\lim_{n \to \infty} F(n) - \varphi^{n+1} = \lim_{n \to \infty} \varphi^n \left( -\frac 1 2 - \frac{3}{2 \sqrt 5} \right)$$

The coefficient of the golden ratio here is clearly less than zero - about $-1.1708$ to be precise. Meanwhile, $\varphi \approx 1.618 > 1$, so if $n \to \infty$, then $\varphi^n \to \infty$. Thus,

$$\lim_{n \to \infty} F(n) - \varphi^{n+1} = -\infty$$

Thus, we've found that

$$L = 2^{\lim \limits_{n \to \infty} F(n) - \varphi^{n+1}} = 0$$

since the limit in the exponent is $-\infty$. This intuitively makes sense. Even at $n=10$, $f(n)/g(n) = 2^{F(n) - \varphi^{n+1}} \approx 10^{-44}$ per Wolfram.

PrincessEev
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