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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!

Leorio
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  • Hint (for the second one, though the first is similar): let $x$ be the answer. What is $x^2-1$? – lulu Jan 14 '18 at 21:31
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    He was half telling the truth. You need advanced math to know that any of the expressions actually equal anything. – fleablood Jan 14 '18 at 21:31
  • But notice if $1+{1\over 1+{1\over 1+{1\over 1+{1\over 1+...}}}} = x$ then $1+{1\over 1+{1\over 1+{1\over 1+{1\over 1+...}}}}= 1+\frac 1x$. So $x = 1 + \frac 1x$ so...... – fleablood Jan 14 '18 at 21:33
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    The question about nested radicals has been answered several times in the past (see the above linked question, or this question). The continued fraction has also been dealt with before (for example, here and here, and here). – Xander Henderson Jan 14 '18 at 21:40

1 Answers1

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$\bullet$ 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 ..

Atmos
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