$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation

$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=7^\frac{1}{2}\cdot7^\frac{1}{4}\cdot 7^\frac{1}{8}\cdots=7^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots}=7^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}=7$$

Your expression can be written as $$7^{\frac12 + \frac14 ...}.$$

Now you can use sum of infinite GP = $\frac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio.

Thus sum $= 1$.

Your expression $=$ $7^1$ = $7$

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ If $\exists\ \lim_{n \to \infty}x_{n} = s > 0$: $$ s = \root{7s}\quad\imp\quad s = 7 $$ Also $$ x_{n} - 7 = \root{7}x_{n - 1}^{1/2} - 7 = {7x_{n - 1} - 49 \over \root{7}x_{n - 1}^{1/2} + 7} ={x_{n - 1} - 7 \over \root{x_{n - 1}/7} + 1} < x_{n - 1} - 7 $$