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

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It would probably be easier to think of it if you write it as a fractional exponent: $7^\frac{1}{32}$ –  Dan the Man Dec 2 '13 at 14:57
    
Actually, in general, root(n*x) is closer to n than x. Unless n is x. From that, if you do it an infinite number of times, x will always get to n –  Cruncher Dec 2 '13 at 18:06
    
"The square root approximation". Good episode title for Big Bang Theory! –  Kaz Dec 3 '13 at 2:01
    
@user2378, please stop adding the roots tag. It has nothing to do with this question. Take a look at the tag's description. –  Antonio Vargas Dec 3 '13 at 23:59

6 Answers 6

up vote 69 down vote accepted

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

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2  
$x^2-7x =0$ and hence x=7? –  user2378 Dec 2 '13 at 7:25
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You didnt prove convergence –  Abdulh Khazzak Gustav ElFakiri Dec 2 '13 at 12:50
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@AbdulhKhazzakGustavElFakiri, we can apply the logic heer (math.stackexchange.com/questions/324650/…) $$\sum_{n = 0}^\infty 7^{\frac1{2^n}}=\ln 7 \sum_{n = 0}^\infty \frac1{2^n}=\cdots$$ –  lab bhattacharjee Dec 2 '13 at 15:12
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Down voter, would you mind disclosing the mistake? –  lab bhattacharjee Dec 4 '13 at 15:52
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@HarshalGajjar, Thanks. But, I implore you not to help. –  lab bhattacharjee Sep 20 at 5:04
up vote 102 down vote
+100

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

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6  
this hints looks so fresh!! I usually do this by Mr.Labbhattacharjee's way... –  Praphulla Koushik Dec 2 '13 at 7:25
    
Ah!! I am disappointed... This looks so nice if it was left just by writing $\sqrt{7\sqrt{7\sqrt{7\dots}}}=7^{\frac{1}{2}}7^{\frac{1}{4}}7^{\frac{1}{8}}...$ –  Praphulla Koushik Dec 2 '13 at 7:27
    
Sorry sir, but exactly what you say –  Madrit Zhaku Dec 2 '13 at 7:28
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it would have been look much great if you have removed $=7^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots}=7^{\frac{\frac{1}{2}}{1-\frac{1‌​}{2}}}=7$ part... –  Praphulla Koushik Dec 2 '13 at 7:31
    
what did help was the beginning, and this is now the complete solution of the example –  Madrit Zhaku Dec 2 '13 at 7:33

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 $=$ $7^1$ = $7$

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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, $y=7$. $$\displaystyle \boxed{\therefore \sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=7}$$

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

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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 $$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{\cdots}}}}}=\large\lim_{n\to\infty}\, a_n=\lim_{n\to\infty}\,7^{1-\Large2^{-n}}=\bbox[3pt,border:3px #FF69B4 solid]{\color{red}{7}}$$

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