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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 $Q$?

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    $\begingroup$ Did you mean $(x!)^{2^{-x}}\to 1?$ That's the usual test (Herschfeld's Convergence Theorem) for whether the nested radical converges. The limit you wrote actually exists, and is zero. $\endgroup$ – Adrian Keister Dec 12 '19 at 17:57
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    $\begingroup$ Interesting that you can set up a recurrence relation. If you let $Q_n=\sqrt{n!+\sqrt{(n+1)!+\dots}},$ then $Q_{n+1}=Q_n^2-n!.$ Don't know if that helps you or not - Mathematica can't solve it. $\endgroup$ – Adrian Keister Dec 12 '19 at 18:05
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    $\begingroup$ @AdrianKeister Yes, edited. $\endgroup$ – Descartes Before the Horse Dec 12 '19 at 18:16
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    $\begingroup$ First, it appears that your $Q \approx 1.827014718$ while the result from the inverse symbolic calculator is for $1.827014766$; it seems unlikely that these two are actually the same. Second, $sr(5)$ in the inverse symbolic calculator (wayback.cecm.sfu.ca/projects/ISC/ISCmain.html) is actually the notation for the square root of 5. In fact $(e^{\pi} / \sin(\pi/12))^\sqrt{5} = 0.01827014\cdots$ - the ISC also ignores powers of 10. $\endgroup$ – Michael Lugo Dec 12 '19 at 19:14
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    $\begingroup$ There is a possibly relevant Quora post: (Link) which includes a proof that $\sqrt{2!^2+2!\sqrt{3!^2+3!\sqrt{\ldots}}}=8$. The constant also has the OEIS sequence (A099876) but given that it doesn't lead anywhere fruitful, I think the problem's likely going to remain open. $\endgroup$ – Jam Dec 12 '19 at 21:50
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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.

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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 $1.29970\times10^{-7}$

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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 !

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