Finding the value of $\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$ 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!
 A: 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:
$$
a_5 \approx 3.079604451993
\\
a_6 \approx 3.081712705722
\\
a_7 \approx 3.082633123037
\\
a_8 \approx 3.083036100688
\\
a_9 \approx 3.083213386604 
\\
a_{10} \approx 3.083291812809
$$
A: 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.
A: 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 $<10^{-100}$.
A: 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 $p=\frac{(\sqrt{1+2\sqrt{2}}-1)+1}{0.4755} \approx 3.114$
A: 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.
