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This question already has an answer here:

What is the sum of this infinite series of roots: $$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4 + \cdots+\sqrt{\infty}}}}}$$

This is an interesting expression because the increase created by the addition of the next term is offset by the square root introduced with the next term. the key is determining what the relationship between these two is exactly

I know it converges on an irrational number around 1.75 as I have tried computing it for a few small iterations. But I was wondering how to algebraically simplify this expression.

I have tried many things and ways to manipulate the expression but to no avail.

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marked as duplicate by Chris Culter, Thomas Andrews, Nick Peterson, Amitesh Datta, Stahl Jan 10 '14 at 6:21

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    $\begingroup$ Can you prove that the expression actually converges? $\endgroup$ – Igor Rivin Jan 10 '14 at 4:19
  • $\begingroup$ no i cant, i only assumed $\endgroup$ – Rizzy C Jan 10 '14 at 4:26
  • $\begingroup$ To show convergence, see: mathworld.wolfram.com/HerschfeldsConvergenceTheorem.html. But after that, how do you even know that the number is irrational? $\endgroup$ – Nick D. Jan 10 '14 at 4:26
  • $\begingroup$ Actually, it is not known whether this number is irrational. $\endgroup$ – Julien Godawatta Jan 10 '14 at 4:34
  • $\begingroup$ The $\large\infty$ in the formula avoids to write $\large n$ and put in from of it the $\large\lim_{n \to \infty}$. It's clearly understood, by people who know have 'to read between lines' ,that is a short cut. You can leave it there. $\endgroup$ – Felix Marin Jan 10 '14 at 4:37
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This is the Nested Radical Constant , which does converge, but whose closed-form expression is still unknown. See also Somos's Quadratic Recurrence Constant for a similar constant, involving their product.

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Hint: $$ a_{1} = \root{1 + \root{a_{2}}}\,,\quad a_{2} = \root{2 + \root{a_{3}}}\,, \ldots\,, a_{n} = \root{n + \root{a_{n + 1}}} $$ Then, $\ds{a_{n + 1} = \pars{a_{n}^{2} - n}}^{2}$. $\tt\mbox{Calculate}$ $a_{1}$.

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    $\begingroup$ How is this a hint? $\endgroup$ – Did Jan 10 '14 at 22:54

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