Prove that $1+{1\over 1+{1\over 1+{1\over 1+{1\over 1+...}}}}=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}$

Question

So, my professor me gave this exercise as a challenge:

-First, prove that:

$$1+{1\over 1+{1\over 1+{1\over 1+{1\over 1+...}}}}={1+\sqrt{5}\over 2}.$$

-Then, prove that:

$$1+{1\over 1+{1\over 1+{1\over 1+{1\over 1+...}}}}=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}$$

He said you need no advanced maths to solve it: you just need high-school math with no calculus...

The thing is that it's been 2 weeks now and I'm as lost as when I first saw the problem.

Can someone help me!

Answer 1

$\bullet$ Consider the sequence given by $$u_{n+1}=1+\frac{1}{u_n}$$ with $u_0=1$

You can show that this sequence is convergent and positive, and hence converges to a real that satisfies $$ \ell=1+\frac{1}{\ell} \Leftrightarrow \ell^2-\ell-1=0 $$ You 'll find that it has two potential solutions but only one positive you'll find that

$$\displaystyle \ell=\frac{1+\sqrt{5}}{2}$$

And

$$u_{n+1}=1+\frac{1}{u_n}=1+\frac{1}{1+\displaystyle\frac{1}{u_{n-1}}}=1+\frac{1}{1+\displaystyle\frac{1}{1+\displaystyle \frac{1}{u_{n-2}}}}$$

Etc etc .... So this is an approach of the first "equality", but you need continued fraction knowledge to really understand and prove the equality.

$\bullet$You can then study $v_n$ given by $v_0=1$ and $$ v_{n+1}=\sqrt{1+v_n} $$ You can show that this sequence is increasing and bounded and with the same idea $$ \ell=\sqrt{1+\ell} \Leftrightarrow \ell^2=1+\ell $$ which by positivity will lead you to $\ell$ again. And

$$v_{n+1}=\sqrt{1+v_n}=\sqrt{1+\sqrt{1+v_n}}=\sqrt{1+\sqrt{1+\sqrt{1+v_{n-1}}}}$$

Etc etc ..