Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
It would probably be easier to think of it if you write it as a fractional exponent: $7^\frac{1}{32}$ – Daniel Pendergast 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
up vote 76 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

share|cite|improve this answer
$x^2-7x =0$ and hence x=7? – user2378 Dec 2 '13 at 7:25
You didnt prove convergence – Abdulh Khazzak Gustav ElFakiri Dec 2 '13 at 12:50
@AbdulhKhazzakGustavElFakiri, we can apply the logic heer (…) $$\sum_{n = 0}^\infty 7^{\frac1{2^n}}=\ln 7 \sum_{n = 0}^\infty \frac1{2^n}=\cdots$$ – lab bhattacharjee Dec 2 '13 at 15:12
Down voter, would you mind disclosing the mistake? – lab bhattacharjee Dec 4 '13 at 15:52
@HarshalGajjar, Thanks. But, I implore you not to help. – lab bhattacharjee Sep 20 '14 at 5:04
up vote 132 down vote

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

share|cite|improve this answer
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
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 $=$ $7^1$ = $7$

share|cite|improve this answer

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

share|cite|improve this answer

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

share|cite|improve this answer

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

share|cite|improve this answer

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

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

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

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

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

share|cite|improve this answer
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. – Steven Stadnicki Feb 5 '15 at 18:43
I did an experiment. $\sqrt(7) < \sqrt{7\sqrt{7}} < \sqrt{7\sqrt{7\sqrt{7}}}$ – Ama Feb 5 '15 at 19:24
$$\sqrt{n} < \sqrt{n \sqrt{n}}$$ – Ama Feb 5 '15 at 19:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.