# How to evaluate $\sqrt {6\sqrt {6\sqrt{\cdots}}}$ [duplicate]

I was playing mental maths in which we solve problems without pen and pencil.

Then I came across a problem $$\sqrt {6 \sqrt{6 \sqrt{6\sqrt\cdots}}}.$$ I asked my teacher and he told me that the answer is $6$

Kindly tell me what was the reason behind my teacher's saying...

I am still amazed!!!!

## marked as duplicate by Hans Lundmark, user1551, Dietrich Burde, saz, NamasteDec 9 '16 at 0:54

• Worth checking this math.stackexchange.com/questions/115501/… – Steve Dec 8 '16 at 8:25
• This is not a full proof but if $x=\sqrt{6\sqrt{6\sqrt{6...}}}$ then $x = \sqrt{6x}$. – Levent Dec 8 '16 at 8:25
• I am not familiar with convergence of series...!!!! Sorry – Atul Mishra Dec 8 '16 at 8:27
• Assume the expression to be equal to x. Then square both sides.u will find a quadratic equation like x^2=6x and solve it...... – Sujan Dutta Dec 8 '16 at 8:27

Suppose that expression indeed defines a number, call it $x$: $$x = \sqrt{6\sqrt{6\sqrt{6\sqrt{\cdots}}}}$$ Square both sides: $$x^2 = 6\color{blue}{\sqrt{6\sqrt{6\sqrt{\cdots}}}}$$ Notice that the blue part is $x$ again, so: $$x^2 = 6x$$ This equation has two solutions; but one of them is...

From the comment:

why $x \ne 0$ ?

Take a look at the sequence: $$\sqrt{6}, \sqrt{6\sqrt{6}},\sqrt{6\sqrt{6\sqrt{6}}}, \ldots$$ This sequence is increasing with first term $\sqrt{6}>0$ so if this converges, it cannot converge to $0$.

To show this increasing sequence converges, you only need that it is bounded. Call the $n$-th term in the sequence above $x_n$ and observe that $x_1 = \sqrt{6} \le 6$. Now if $x_n \le 6$, then by induction also $x_{n+1} = \sqrt{6x_n} \le \sqrt{6 \cdot 6} = 6$. Thanks to Bungo for his comments.

You can also take a look at this similar question.

• why $x \neq 0$ ??? – Atul Mishra Dec 8 '16 at 8:29
• See addition in the answer. – StackTD Dec 8 '16 at 8:31
• Indeed; I added a reference concerning the boundedness. – StackTD Dec 8 '16 at 8:37
• I will add this; thank you for the suggestion. – StackTD Dec 8 '16 at 8:51
• (+1) Nice, complete answer which includes the convergence proof, without which the $x^2 = \sqrt{6x}$ is unjustified. I deleted my previous comments now that you have integrated them into your answer. – Bungo Dec 8 '16 at 8:54

Another possible solution: just compute the geometric sum in the exponent.

$$x=6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...}=6$$

The first embedded infinite radical is identical to the whole thing. Supposing that the whole thing is $6$, the nested infinite radical is also $6$. When we multiply this by the leftmost $6$, we get $\sqrt{36}$, which is $6$, meaning it holds.

Hint Consider the function $f(x)=\sqrt{6x}$ and evaluate the limit of the sequence given by the recurrence $$x_1=1,\quad x_{n+1}=f(x_n),\quad\mbox{for n\geq 1}$$ Show that $x_n$ is increasing and bounded and it tends to $L$ where $f(L)=L$.

It is important that you look for a possible recursion in these questions.

Suppose $x=\sqrt{6\sqrt{6\sqrt{6\sqrt{6\ldots}}}}$. Then, see that $x = \sqrt{6x}$ (Can you see why? Basically, what was happening outside the radical is now happening inside the radical).

Now, square it so that $x^2 = 6x$ hence $x=6$ or $x=0$.

To show that $x \neq 0$, do the following: define $x_1 = \sqrt{6}$, and let $x_{n+1} = \sqrt{6x_n}$. By induction prove ,that $x_n$ is an increasing strictly positive sequence. Then prove that $x = \lim x_n$, hence $x$ also must be positive, hence can't be zero.

As another example, you can consider $x= \frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}$. This also can be rewritten recursively, to give $x = \frac{1}{1+x}$, which when solved gives $x = \frac {\sqrt 5 - 1}2$. Hence, $$\frac {\sqrt 5 - 1}2 = \frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}$$

There are many more of these marvels. Try to create some on your own.

If you start with $$x = \sqrt{6\sqrt{6\sqrt{6\cdots}}}$$ with an infinite number of terms then $$x=\sqrt{6x}$$ which you can solve. Depending on how you do it, perhaps by squaring both sides, this could give two potential solutions and you need to satisfy youself about which, if any, is correct.

assume that $\sqrt {6 \sqrt{6 \sqrt{6\sqrt....}}}=x$

$x^2=6\times \sqrt {6 \sqrt{6 \sqrt{6\sqrt....}}}$

$x^2=6\times x$

$x=6$ Or $x=0$.

But Notice that $x=\sqrt{6}\times \sqrt{6}^{1/2}\times\sqrt{6}^{1/4}......$

$x=\sqrt{6}^{{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...}}$.The series on the right is never gonna converge to zero. I shall let you conclude now.

• Here x could also be 0,, why it is not?? – Atul Mishra Dec 8 '16 at 8:28