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Is it possible to find the value of $$\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$ Does it help if I set it equal to $x$? Or I mean what can I possibly do? $$x=\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$ $$x^2=1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$ $$x^2-1=2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$ $$\frac{x^2-1}{2}=\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$ $$\left(\frac{x^2-1}{2}\right)^2=2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}$$ $$\left(\frac{x^2-1}{2}\right)^2-2=3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}$$ $$\vdots$$

I don't see it's going anywhere. Help appreciated!

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    $\begingroup$ Numerically, it seems to converge pretty quickly to $3.08335514183069\ldots$ The Inverse Symbolic Calculator finds nothing. $\endgroup$ – Rahul Jul 4 '13 at 3:40
  • $\begingroup$ If you chop it off at 1, the answer is 1. If you chop it off at 2, the answer is $\sqrt{1+2\sqrt{2}}$ or nearly 2. Carry on like that for a few terms. Even the first ten terms shouldn't take too long on a calculator, and you can see whether it is levelling off, and if it is, you can get an estimate of the limit. EDIT I wrote that before Rahul's comment. $\endgroup$ – Empy2 Jul 4 '13 at 3:41
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    $\begingroup$ From your first line $x=$ to the second you should have $x^2=$. I recognize that this makes it harder.... $\endgroup$ – Ross Millikan Jul 4 '13 at 3:49
  • $\begingroup$ @Mahmud The mighty Alpha doesn't even interpret the form correct... $\endgroup$ – Shuhao Cao Jul 4 '13 at 4:07
  • $\begingroup$ Thanks, I deleted it @ShuhaoCao ; but I really want to know the answer! :) $\endgroup$ – Sniper Clown Jul 4 '13 at 4:09
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This is meant to follow up on Ethan's comment about using Herschfeld's theorem to prove that the expression converges.

Theorem (Herschfeld, 1935). The sequence $$ u_n = \sqrt{a_1 + \sqrt{a_2 + \cdots + \sqrt{a_n}}} $$ converges if and only if $$ \limsup_{n\to\infty} a_n^{2^{-n}} < \infty. $$

The American Mathematical Monthly, Vol. 42, No. 7 (Aug-Sep 1935), 419-429.

In our case we have

$$ \begin{align} u_1 &= \sqrt{1}, \\ u_2 &= \sqrt{1 + 2\sqrt{2}} = \sqrt{1 + \sqrt{2^3}}, \\ u_3 &= \sqrt{1 + 2\sqrt{2 + 3\sqrt{3}}} = \sqrt{1 + \sqrt{2^3 + \sqrt{2^4 3^3}}}, \\ u_4 &= \sqrt{1 + \sqrt{2^3 + \sqrt{2^4 3^3 + \sqrt{2^8 3^4 4^3}}}}, \end{align} $$

and so on, so that

$$ a_n = n^3 \prod_{k=2}^{n-1} k^{2^{n-k+1}}. $$

We then have

$$ a_n^{2^{-n}} = n^{3\cdot 2^{-n}} \prod_{k=2}^{n-1} k^{1/2^{k-1}} \sim \prod_{k=2}^{\infty} k^{1/2^{k-1}} $$

as $n \to \infty$, where the infinite product converges because $k^{1/2^{k-1}} = 1 + O(\log k/2^k)$ as $k \to \infty$. Therefore $u_n$ converges by Herschfeld's theorem.

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Not an answer of a closed form, but we can use Ramanujan's formula to approximate:

For any $n\in \mathbb{N}$ $$f(n) = 1+ n = \sqrt{1+n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)\sqrt{1+\dots}}}}}.$$

Let the sum be $x$.

$$ \begin{aligned} x &> \sqrt{1+2\sqrt{\color{blue}{1}+3\sqrt{\color{blue}{1}+4\sqrt{\color{blue}{1}+5\sqrt{\color{blue}{1}+\dots}}}}} \\&= \sqrt{1+2f(3)}:=a_1 = 3. \\ x &> \sqrt{1+2\sqrt{2+3\sqrt{\color{blue}{1}+4\sqrt{\color{blue}{1}+5\sqrt{\color{blue}{1}+\dots}}}}} \\&= \sqrt{1+2\sqrt{2+3f(4)}} :=a_2\approx 3.040758335 \\ x &> \sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{\color{blue}{1}+5\sqrt{\color{blue}{1}+\dots}}}}} \\&= \sqrt{1+2\sqrt{2+3\sqrt{3+4f(5)}}} :=a_3\approx 3.063938469 \\ x &> \sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{\color{blue}{1}+\dots}}}}} \\&= \sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5f(6)}}}} :=a_4\approx 3.074786007. \end{aligned} $$ So the nested radical does not go on infinitely in this sequence $\{a_n\}$, where $$ a_n = \sqrt{1+2\sqrt{2+3\sqrt{3+\dots\sqrt{\dots\sqrt{(n-1)+n\sqrt{n+(n+1)(n+3)}}}}}}. $$ And if we compute a few more terms: $$ a_5 \approx 3.079604451993 \\ a_6 \approx 3.081712705722 \\ a_7 \approx 3.082633123037 \\ a_8 \approx 3.083036100688 \\ a_9 \approx 3.083213386604 \\ a_{10} \approx 3.083291812809 $$

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In the spirit of Ramanujan, let

$$G(x)=\sqrt{x-1+x\sqrt{(x+n-1)+(x+n)\sqrt{(x+2n-1)+(x+2n)\sqrt{\ldots}}}}$$

and note that by setting $x=2$ and $n=1$ we recover our nested radical. After squaring,

$$G(x)^{2}=(x-1)+xG(x+n)$$

This is a slightly more complicated functional than Ramanujan encountered and I'm not sure how to solve it, but hopefully someone else can provide more insight.

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We can write $$ \begin{align} x & = \sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\cdots}}}}} \\ & = \sqrt{1+2\sqrt{2}\sqrt{1+\frac{3\sqrt{3}}{2}\sqrt{1+\frac{4\sqrt{4}}{3}\sqrt{1+\frac{5\sqrt{5}}{4}\sqrt{1+\cdots}}}}} \\ &< \sqrt{1+2\sqrt{2}\sqrt{1+\frac{3\sqrt{3}}{2}\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots}}}}}} \end{align} $$ since $$ \frac{n\sqrt{n}}{n-1}<n-1 \quad \mathrm{for~n>3} $$ Then using the Ramanujan identity @ShuhaoCao cited $$ 1+n = \sqrt{1+n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+\cdots}}}} $$ it follows that $x$ converges and we can get upper bounds $$ x < \sqrt{1+2\sqrt{2}\sqrt{1+\frac{3\sqrt{3}}{2}\cdot (1+3)}} = 3.247\cdots \\ x< \sqrt{1+2\sqrt{2}\sqrt{1+\frac{3\sqrt{3}}{2}\sqrt{1+\frac{4\sqrt{4}}{3}\sqrt{1+\frac{5\sqrt{5}}{4}\cdot (1+5)}}}} = 3.159\cdots $$ and so forth.

Refining these upper bounds and the lower bounds described by @ShuhaoCao $$ a_n = \sqrt{1+2\sqrt{2+3\sqrt{3+\cdots\sqrt{n+(n+1)(n+3)}}}} \\ b_n = \sqrt{1+2\sqrt{2}\sqrt{1+3\sqrt{3}\sqrt{1+\cdots\sqrt{1+\frac{(n+1)^{3/2}}{n}(n+2)}}}} \\ a_n<x<b_n \quad n>1 $$ by computing $a_{1000},b_{1000}$ we can find $$ x = 3.083355141830694458051142580088\cdots $$ with a range of $<10^{-100}$.

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Let $p=\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$
Define :$$x_1=\sqrt{1+2\sqrt{2}}-1$$ $$x_2=\sqrt{1+2\sqrt{2+\sqrt{3}}}-\sqrt{1+2\sqrt{2}}$$ $$.$$ $$.$$ $$x_{n-1}=\sqrt{1+2\sqrt{2+\sqrt{3+...+\sqrt{n}}}}-\sqrt{1+2\sqrt{2+\sqrt{3+...+\sqrt{n-1}}}}$$ From the summation , we can see that $$p-1=x_1+x_2+....+x_{n-1}$$ $$x_1 \approx 0.956636$$ $$x_2 \approx 0.566284$$ $$x_3 \approx 0.290212$$ $$x_4 \approx 0.141296$$ $$x_5 \approx 0.067556$$ not kosher, but ratio approximation will be $0.4755$ ,then $p=\frac{(\sqrt{1+2\sqrt{2}}-1)+1}{0.4755} \approx 3.114$

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