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 $6$
I was amazed and asked how but didn't get satisfying reply.
Kindly tell me what was the reason behind my teacher's saying...
If this question is asked somewhere else, please send the link in comments
I am still amazed!!!!
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.
Another possible solution: just compute the geometric sum in the exponent.
$$x=6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...}=6$$
Could be done in your head if you wanted to.
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.
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 $f(L)=L$.
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.
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.
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.
