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!

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.
