Evaluating $Q = \sqrt{1!+\sqrt{2!+\sqrt{3!+\dots}}}$ Consider the nested radical
$$Q = \sqrt{1!+\sqrt{2!+\sqrt{3!+\sqrt{4!+\sqrt{5!+\sqrt{6!+\dots}}}}}}\, .$$
I'm certain the above nested root converges, considering $(x!)^{2^{-x}} \to 1$ (is this a sufficient condition to conclude convergence?) I calculated $Q$ to quite a few digits, and to my surprise, found something that was off by less than $1 \times 10^{-9}$:
$$A=\left(\frac{e^{-\pi}}{\sin(\frac\pi{12})}\right)^{\sqrt{5}}.$$
Which isn't pretty, but it is very close to $Q$, and may be even closer (or, less excitingly, less close) if I calculated $Q$ to more digits. Unfortunately, $Q$ grows insanely slowly, so I can't really do that. Whether this actually converges to this strange number or not, is there any hope in finding a closed form for $Q$?
 A: Just for the fun of it !
I gave a friend of mine the constant as printed in sequence $A099876$ at $OEIS$ and he came back with the approximation
$$\frac{2-\sqrt{2}+2 \sqrt{3}-3 e-2 \pi -5 \pi ^2+\log (2)+2 \log (3)}{2 \sqrt{2}+8
   \sqrt{3}-9 e+8 \pi -6 \pi ^2+6 \log (2)+6 \log (3)}$$ which is in a relative error of $8.14 \times 10^{-18}$%.
Not very nice !
A: This is nothing more than a long comment, but I found myself wondering if there was an easy way to get a reasonable upper bound on the value of $Q$, assuming its limit exists. (Lower bounds are a dime a dozen; any truncation of the nested radical will do.) Here's what I came up with:
$$\begin{align}
\sqrt2Q&=
\sqrt2\sqrt{1!+\sqrt{2!+\sqrt{3!+\cdots}}}\\
&=\sqrt{2+2\sqrt{3!+\sqrt{4!+\cdots}}}\\
&=\sqrt{2+\sqrt{4\cdot2!+4\sqrt{3!+\cdots}}}\\
&=\sqrt{2+\sqrt{4\cdot2!+\sqrt{16\cdot3!+\sqrt{256\cdot4!+\cdots}}}}\\
&\gt\sqrt{2+\sqrt{3!+\sqrt{4!+\sqrt{5!+\cdots}}}}\\
&=Q^2-1
\end{align}$$
so $Q^2-\sqrt2Q-1\lt0$, which implies
$$Q\lt{\sqrt2+\sqrt6\over2}\approx1.93$$
This bound, while crude because crudely obtained, is nonetheless not too far from the reported value, $Q\approx1.827$. 
Added later: It seems worth giving a proof that the sequence $Q_n=\sqrt{1!+\sqrt{2!+\sqrt{3!+\cdots+\sqrt n!}}}$ converges. 
It's clear that the sequence is monotonically increasing, so it suffices to show it's bounded above. The proof is by induction (on $n$) of the following statement:  For all $m,n\in\mathbb{N}$, 
$$\sqrt{m!+\sqrt{(m+1)!+\cdots+\sqrt{(m+n)!}}}\le m!+1$$
The inequality is certainly for all $m$ in the base case $n=0$: $\sqrt{m!}\le m!+1$. Induction now says that
$$\sqrt{m!+\sqrt{(m+1)!+\cdots+\sqrt{(m+n)!}}}\le\sqrt{m!+((m+1)!+1)}$$
so it's enough to check that
$$m!+(m+1)!+1\le(m!+1)^2$$
which is easy enough to see, since
$$(m!+1)^2=m!m!+2m!+1=m!+(m!+1)m!+1\ge m!+(m+1)m!+1=m!+(m+1)!+1$$
Letting $m=1$ in the inequality $\sqrt{m!+\sqrt{(m+1)!+\cdots+\sqrt{(m+n)!}}}\le m!+1$, follows that $Q_n\le1!+1=2$ for all $n$, so the (monotonically increasing) sequence is bounded above, hence converges to a limit.
A: We can actually improve the bounds for $Q$, by knowing two basic identities of nested radicals $$\phi=\sqrt{1+\sqrt{1+\sqrt{1+...}}}$$ $$\phi\sqrt{n}=\sqrt{n+\sqrt{n^2+\sqrt{n^4+...}}}$$ The latter is formed by pushing in though $n$ inside the radicals. Now take $$Q^2-1=\sqrt{2+\sqrt{6+\sqrt{24+...}}}$$
Now to get a bound we need to use a nested radical with the starting co-effectients similar to that of $Q$. Take;
$$\phi \sqrt{2^{7/6}}=\sqrt{2^{7/6}+\sqrt{2^{7/3}+\sqrt{2^{14/3}+...}}}$$ $$=\sqrt{2.2449..+\sqrt{5.039684+\sqrt{25.39841..+...}}}$$ It can be seen that $$\phi \sqrt{2^{7/6}}>Q^2-1$$ Subsequently, $$Q<\sqrt{\phi \sqrt{2^{7/6}}+1}$$ $$Q<1.85048960..$$ Not a bad bound considering $Q=1.827014717..$. We can also take this method to a much extend and obtain, $$Q<\sqrt{\phi \sqrt[^8]{4!}+1}$$ $$Q<1.84586304..$$
Another derivable bound would be; $$Q<\sqrt{1+\sqrt{2+\phi \sqrt[^{16}]{15\cdot2^{15}}}}$$ $$Q<1.838818182...$$
An approximation for Q would be 
$$Q\approx e^{W\left(\frac{10}{9}-\frac{1}{100}\right)}$$
Where $W(x)$ is the Lambert W-function, also called the product-log. With the error being $1.29970\times10^{-7}$
