# On proving an infinite-nested radical

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.

Thanks.

Edit:

@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 $$n>0$$.

$$2n+1=\sqrt{2(n+1)(2n-1)+\sqrt{2(n+2)(2n+1)+\sqrt{2(n+3)(2n+3)+\cdots}}}$$
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. $$\;\bigcirc$$