Prove that $ 3 =\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{...}}}}$
According to the source, this can be done easily like
$$(x+1)^2 = 1+2x+x^2$$
$$(x+1)^2 = 1+x(x+2)$$
But how to continue and prove this?
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There is a reasonably accessible explanation of this at https://www.youtube.com/watch?v=leFep9yt3JY or possibly https://mindyourdecisions.com/blog/2016/05/01/ramanujans-nested-radical-sunday-puzzle/ – sjb-2812 Aug 19 '19 at 08:20
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Since $n^2 = 1+(n-1)(n+1)$, then $n = \sqrt{1+(n-1)(n+1)}$. Repeating the idea with $(n+1)^2 = 1+(n)(n+2)$, yields $n= \sqrt{1+(n-1)\sqrt{1+n(n+2)}}$. Further applications result in the infinite expression $n=\sqrt{1+(n-1)\sqrt{1+n\sqrt{1+(n+1)\sqrt{1+\cdots}}}}$.
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1Yes, so...? Even as a hint, that is way too vague and incomplete, imo. – DonAntonio Aug 19 '19 at 08:52
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@DonAntonio How is that vague? It’s not even just a hint; it’s the answer. Just let $n=3$ – Aug 21 '19 at 01:12