Solve $\dfrac{1}{1+\frac{1}{1+\ddots}}$

Question

I'm currently a high school junior enrolling in AP Calculus, I found this website that's full of "math geeks" and I hope you can give me some clues on how to solve this problem. I'm pretty desperate for this since I'm only about $0.4%$ to an A- and I can't really afford a B now...

The problem is to simplify: $$\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}}$$

What I did, was using basic "limits" taught in class and I figured out that the denominator would just keep going like this and approaches $1$, so this whole thing equals $1$, but I think it's not that easy...

Answer 1

Hint: If $a = \frac{1}{1+\frac{1}{1+\cdots}}$ then $\frac{1}{a}-1 = \frac{1}{1+\frac{1}{1+\cdots}} = a$. Can you take it from here?

Answer 2

Let us denote the number given by $x$. Then what is $\dfrac1{1+x}$? How does this help us?

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You may take interest in reading on continued fractions. There's a nice theory about them.

Answer 3

First hint: It’s $1/φ$. Check out the golden ratio.

Second hint, which adds a bit to answers given so far: Since you should probably apply limits formally, you can describe it as the limit of a sequence given by $a_0 = 1$ and $a_{k+1} = \frac{1}{1 + a_k}$. Now if $x = \lim_{k → ∞} a_k$, then $$\frac{1}{1+x} \overset{\text{limit rules}}{=} \lim_{k → ∞} \frac{1}{1 + a_{k}} = \lim_{k → ∞} a_{k+1} = x$$ So $x^2 + x - 1 = 0$.

To show that limit exits in the first place, you can tell from the frist few members: $$ 1,\; \frac{1}{2},\; \frac{2}{3},\; \frac{3}{5},\; \frac{5}{8},\; \frac{8}{13},\; …$$ that the sequence is actually given by $$ a_k = \frac{f_k}{f_{k+1}};\quad \text{where}\; f_{k+2} = f_{k+1} + f_k, \text{and}\; f_1 =f_2 = 1,$$ meaning $f_k$ denotes the $k$-th Fibonacci number, and readily prove this by induction.

Now, the $k$-th Fibonacci number is given by $f_k = \frac{φ^k - ψ^k}{\sqrt{5}}$ where $φ = \frac{1+\sqrt{5}}{2}$ and $ψ = \frac{1 - \sqrt{5}}{2}$, the roots of $X^2 - X - 1$. This is true since the Fibonacci sequence is uniquely determined by the recursion and its first two members. But that foregoing condition implies $φ^{k+2} = φ^{k+1} + φ^k$ as well as $ψ^{k+2} = ψ^{k+1} + ψ^k$, so: $$\frac{φ^{k+2} - ψ^{k+2}}{\sqrt{5}} = \frac{φ^{k+1} - ψ^{k+1}}{\sqrt{5}} + \frac{φ^k - ψ^k}{\sqrt{5}},$$ whereas $\frac{φ^1 - ψ^1}{\sqrt{5}} = \frac{φ^2 - ψ^2}{\sqrt{5}} = 1$.

Now $\frac{f_k}{φ^k} \overset{k → ∞}{\longrightarrow} \frac{1}{\sqrt{5}}$, since $0 < ψ < 1$ and so $ψ^k \overset{k → ∞}{\longrightarrow} 0$ and $0 < \frac{1}{φ} < 1$ and so $\frac{1}{φ^k} \overset{k → ∞}{\longrightarrow} 0$. This, however, implies: $$\frac{f_k}{f_{k+1}} = \frac{1}{φ} · \frac{f_k}{φ^k}·\frac{φ^{k+1}}{f_{k+1}} \overset{k → ∞}{\longrightarrow} \frac{1}{φ}.$$

Answer 4

This is the Golden ratio (also known as $\varphi$) expressed using countinued fraction. This number is solution of the $x^2-x-1=0$ quadratic equotation.
This quadratic equotation you can wrote as $x=1+\frac{1}{x}$ and this form is used to construct cotinued faction used in your question.
See wikipedia for "Golden Ratio" or read book writed by Mario Livio: Golden ratio for example. Here you found more answers.