$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.

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

 A: Let $$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=x $$
Clearly, $x>0$
$$\implies x^2=7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=7x$$
Now left is the proof of converge(as conversed with  Abdulh Khazzak Gustav ElFakiri)
Observe that the $r$th term $T_r$ of this infinite product is $\displaystyle7^{\left(\frac1{2^r}\right)}$
using Convergence/Divergence of infinite product, $$\sum_{0\le r<\infty}\ln(T_r)=\ln 7\sum_{0\le r<\infty}\frac1{2^r}$$ which is an infinite Geometric Series with common ratio $=\frac12$ which $\in(-1,1)$, hence  the later Series is convergent $\left(\text{ in fact }\displaystyle=\ln7\cdot\frac1{1-\frac12}\right)$, so will be the original infinite Product
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
If $\exists\ \lim\limits_{n \to \infty}x_{n} = s > 0$:
$$
s = \root{7s}\quad\imp\quad s = 7
$$
Also
$$
x_{n} - 7 = \root{7}x_{n - 1}^{1/2} - 7
= {7x_{n - 1} - 49 \over \root{7}x_{n - 1}^{1/2} + 7}
={x_{n - 1} - 7 \over \root{x_{n - 1}/7} + 1} < x_{n - 1} - 7
$$
A: 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 $=$ $7^1$ = $7$
A: $$x = \sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
$$x = \sqrt{7x}$$
$$x^2 - 7x = 0$$
$$x(x - 7) =0 \implies x = 7$$
Because $\sqrt{7} > 0$ we reject the $x=0$ solution.
A: $$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7...}}}}}=7^\frac{1}{2}\cdot7^\frac{1}{4}\cdot 7^\frac{1}{8}\cdots=7^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots}=7^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}=7$$
A: We need to find the value of $\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{\dots}}}}}$.
Step 1: Let $\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=y$
Step 2: Square both sides.
$$7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=y^2$$
Step 3: Recall that $\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=y$. So:
$$7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=7y$$
Step 4: Rewrite the equation.
$$7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=y^2$$
$$7y=y^2$$
$$y^2-7y=0$$
Step 5: Solve for $y$.
$$y^2-7y=0$$
$$y(y-7)=0$$
$$y=0, \ 7$$
It is impossible that $y=0$. So, $y=7$.
$$\displaystyle \boxed{\therefore \sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=7}$$
A: Alternatively, let $a_1,\,a_2,\,a_3,\,\cdots,\,a_n$ be the following sequence
$$\sqrt{7},\,\sqrt{7\sqrt{7}},\,\sqrt{7\sqrt{7\sqrt{7}}},\,\cdots,\underbrace{\sqrt{7\sqrt{7\sqrt{7\sqrt{\cdots\sqrt{7}}}}}}_{\large n\,\text{times}}$$
respectively.
Notice that
$$\large a_n=7^{\Large 1-2^{-n}}$$
Hence
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{\cdots}}}}}=\large\lim_{n\to\infty}\, a_n=\lim_{n\to\infty}\,7^{1-\Large2^{-n}}=\bbox[3pt,border:3px #FF69B4 solid]{\color{red}{7}}$$
