I was playing around with square roots and I noticed that the number $1$ can be seemingly expressed as an infinite nested radical with an easy pattern. I then noticed that if this is true, this would mean that each nested radical inside is equal to the next odd number, respectively. I have tried to denote this using LaTeX but it did not display as I thought it might, though I hope you can understand what I am trying to concisely say.

Conjecture: $$\begin{align}\color{green}1 &= \sqrt{2\cdot \color{red}{-1} + \color{green}{3}} \\ &= \sqrt{2\cdot \color{red}{-1} + \sqrt{2\cdot \color{red}2 + \color{green}{5}}} \\ &= \sqrt{2\cdot \color{red}{-1} + \sqrt{2\cdot \color{red}2 + \sqrt{2\cdot \color{red}{9}+\color{green}7}}}\\ &= \sqrt{2\cdot \color{red}{-1} + \sqrt{2\cdot \color{red}2 + \sqrt{2\cdot \color{red}{9}+\sqrt{2\cdot\color{red}{20}+\color{green}9}}}}\\ &= \sqrt{2\cdot \color{red}{-1} + \sqrt{2\cdot \color{red}2 + \sqrt{2\cdot \color{red}{9}+\sqrt{2\cdot\color{red}{20}+\sqrt{2\cdot \color{red}{35}+\cdots}}}}}\end{align}$$ $\color{red}{-1}+(4\cdot 0+3)=\color{red}{2}$ (radical $=\color{green}1$)
$\color{red}2+(4\cdot 1+3)=\color{red}9$ (radical $=\color{green}3$)
$\color{red}9+(4\cdot 2+3)=\color{red}{20}$ (radical $=\color{green}5$)
$\color{red}{20}+(4\cdot 3+3)=\color{red}{35}$ (radical $=\color{green}7$)
$\cdots = \cdots$

Can this be proven? If it is a result of a generalised identity for odd numbers $2n+1$ (for some $n$, this case being $n=0$), please let me know.



@TheSimpliFire kindly gave the recurrence relation: $1=2a_0+b_n$ where $b_n=\sqrt{2a_n+b_{n+1}}$ and $a_n+4n+3=a_{n+1}$ with $a_0=-1$ and $n>0$.


This is the general identity:


Secondary (albeit similar) Proof:

This proof is my original one. @TheSimpliFire's is way more elegant and concise, but since that first proof is in the post, I will post mine here, just to contribute in benefit of Math.SE I suppose.

We will organise the key elements of the data pairwise as thus: $$(1, -1), (3, 2), (5, 9), (7, 20), (9, 35)$$ Odd numbers are generated from $2n+1$, thus this is equivalent to $$(0, -1), (1, 2), (2, 9), (3, 20), (4, 35)$$ Notice: $$\begin{align} 0 &= (-1 + 1)\div 1 \\ 1 &= (2+1)\div 3 \\ 2 &= (9+1)\div 5 \\ 3 &= (20+1)\div 7 \\ 4 &= (35+1)\div 9 \\ \therefore n &= (n(2n+1)-1+1)\div (2n+1)\end{align}$$ Matching this pattern to the form of the conjecture, we get $2n+1=\sqrt{2(n(2n+1)-1)+\cdots}$ such that $n(2n+1)-1=2n^2+n-1=(n+1)(2n-1)$. $$\therefore 2n+1=\sqrt{2(n+1)(2n-1)+\cdots}$$ Since $(2n+1)^2-2(n+1)(2n-1)=2n+3=2(n+1)+1$ then, by letting $f(n)=2n+1$, we obtain that $$f(n)= \sqrt{2(n+1)(2n-1)+f(n+1)}$$ and thus the general identity follows. $\;\bigcirc$

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