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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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  • $\begingroup$ Related : math.stackexchange.com/questions/588414/… $\endgroup$
    – lab bhattacharjee
    Commented Dec 2, 2013 at 8:02
  • $\begingroup$ It would probably be easier to think of it if you write it as a fractional exponent: $7^\frac{1}{32}$ $\endgroup$
    – Daniel
    Commented Dec 2, 2013 at 14:57
  • $\begingroup$ 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 $\endgroup$
    – Cruncher
    Commented Dec 2, 2013 at 18:06
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    $\begingroup$ @user2378, please stop adding the roots tag. It has nothing to do with this question. Take a look at the tag's description. $\endgroup$
    – Antonio Vargas
    Commented Dec 3, 2013 at 23:59
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    $\begingroup$ And now everybody guess what's the limit of $\sqrt{8\sqrt{8\sqrt{8\sqrt{8\sqrt{8\cdots}}}}}$. $\endgroup$
    – principal-ideal-domain
    Commented Jan 17, 2015 at 11:25

7 Answers 7

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

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    $\begingroup$ this hints looks so fresh!! I usually do this by Mr.Labbhattacharjee's way... $\endgroup$
    – user87543
    Commented Dec 2, 2013 at 7:25
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    $\begingroup$ Ah!! I am disappointed... This looks so nice if it was left just by writing $\sqrt{7\sqrt{7\sqrt{7\dots}}}=7^{\frac{1}{2}}7^{\frac{1}{4}}7^{\frac{1}{8}}...$ $\endgroup$
    – user87543
    Commented Dec 2, 2013 at 7:27
  • $\begingroup$ Sorry sir, but exactly what you say $\endgroup$
    – Madrit Zhaku
    Commented Dec 2, 2013 at 7:28
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    $\begingroup$ 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... $\endgroup$
    – user87543
    Commented Dec 2, 2013 at 7:31
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    $\begingroup$ I appreciate your answer! I've seen it here some time ago! :-) +1 $\endgroup$
    – Markus Scheuer
    Commented Dec 25, 2014 at 15:23
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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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    $\begingroup$ $x^2-7x =0$ and hence x=7? $\endgroup$
    – user2378
    Commented Dec 2, 2013 at 7:25
  • $\begingroup$ @user2378, what is the other root of $x^2=7x?$ $\endgroup$
    – lab bhattacharjee
    Commented Dec 2, 2013 at 7:26
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    $\begingroup$ You didnt prove convergence $\endgroup$
    – Abdulh Khazzak Gustav ElFakiri
    Commented Dec 2, 2013 at 12:50
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    $\begingroup$ @AbdulhKhazzakGustavElFakiri, we can apply the logic heer (math.stackexchange.com/questions/324650/…) $$\sum_{n = 0}^\infty 7^{\frac1{2^n}}=\ln 7 \sum_{n = 0}^\infty \frac1{2^n}=\cdots$$ $\endgroup$
    – lab bhattacharjee
    Commented Dec 2, 2013 at 15:12
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    $\begingroup$ Down voter, would you mind disclosing the mistake? $\endgroup$
    – lab bhattacharjee
    Commented Dec 4, 2013 at 15:52
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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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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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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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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$$x = \sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$

$$x = \sqrt{7x}$$

$$x^2 - 7x = 0$$

$$x(x - 7) =0 \implies x = 7$$

Because $\sqrt{7} > 0$ we reject the $x=0$ solution.

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    $\begingroup$ It is very easy to have $\lim_{n\to\infty} x_n=0$ while $x_n\gt 0$ for all $n$; your reasoning for rejecting the $x=0$ solution doesn't work without a better convergence argument. $\endgroup$
    – Steven Stadnicki
    Commented Feb 5, 2015 at 18:43
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    $\begingroup$ I did an experiment. $\sqrt(7) < \sqrt{7\sqrt{7}} < \sqrt{7\sqrt{7\sqrt{7}}}$ $\endgroup$
    – Ama
    Commented Feb 5, 2015 at 19:24
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    $\begingroup$ $$\sqrt{n} < \sqrt{n \sqrt{n}}$$ $\endgroup$
    – Ama
    Commented Feb 5, 2015 at 19:24

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