let $\displaystyle x_1 = \sqrt{2}$
and $\displaystyle x_{n+1} = \sqrt{2x_n}$
Then $\displaystyle x_{n+1}^2=2x_n$
If the limiting value of $\displaystyle x_n$ is $\displaystyle x$, then the limiting value of $\displaystyle x_{n+1}$ is also $\displaystyle x$.
So $\displaystyle x^2=2x$
$\displaystyle x^2-2x=0$
$\displaystyle x(x-2)=0$
So we see that $\displaystyle x=2$
Note that the given sequence is increasing and bounded above (you can show easily by induction on n that each member of the sequence is bounded by 2). Thus the sequence is guaranteed to converge.
Hello, rcs!
$\displaystyle \text{Evaluate: }\;2\sqrt{2\sqrt{2\sqrt{2\hdots}}} $
Let: .$\displaystyle x \;=\;2\sqrt{2\sqrt{2\sqrt{2\hdots}}} $
$\displaystyle \text{Square: }\;x^2 \;=\;2\cdot\underbrace{2\sqrt{2\sqrt{2\sqrt{2\hdot s }}}}_{\text{This is }x} $
We have: .$\displaystyle x^2 \:=\:2x \quad\Rightarrow\quad x^2-2x \:=\:0 \quad\Rightarrow\quad x(x-2) \:=\:0 $
Therefore: .$\displaystyle x \:=\:2$
Soroban - I found a little mistake in each of our arguments. The answer is actually 4.
In my argument, I misread the problem and left off the constant 2 in front. Since constants can be pulled outside limits this is no big deal, and you can just multiply my answer of 2 by 2 to get 4.
In your argument, the right hand side of where you wrote "Square" should actually start with a 4 (since 2^2 = 4).
Here is my corrected solution:
Let $\displaystyle x_1 = 2\sqrt{2}$
and $\displaystyle x_{n+1} = 2\sqrt{x_n}$
Then $\displaystyle x_{n+1}^2=4x_n$
If the limiting value of $\displaystyle x_n$ is $\displaystyle x$, then the limiting value of $\displaystyle x_{n+1}$ is also $\displaystyle x$.
So $\displaystyle x^2=4x$
$\displaystyle x^2-4x=0$
$\displaystyle x(x-4)=0$
So we see that $\displaystyle x=4$
Note that the given sequence is increasing and bounded above (you can show easily by induction on n that each member of the sequence is bounded by 4). Thus the sequence is guaranteed to converge.