The limit of the square root of two times the square root of two times.....

Question

i'd like to know how to prove that following sequence

$$\sqrt{2\sqrt2},\sqrt{2\sqrt{2\sqrt2}},\sqrt{2\sqrt{2\sqrt{2\sqrt2}}},...$$

Is convergent, and that it's limit equals two.

In other words, that $$\sqrt{2\sqrt{2\sqrt{2\sqrt2...}}}=2$$

I imagine it's supposed to be simple and straightfoward, but i cant figure it out. I would be grateful for any help you could give me.

Answer 1

Your sequence could be rewritten as:

$$ \begin{cases} a_0=\sqrt{2}\\ a_{n+1}=\sqrt{2a_n} \end{cases} $$

By induction, you can easily prove that:

$$ a_n=2^{\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}}=2^{(1-\frac{1}{2^n})}, n\geq 1 $$

The sum in the power converges, hence the sequence converges and the limit is $2$

Answer 2

An idea: define

$$x_1:=\sqrt2\;,\;\;x_2=\sqrt{2x_1}\,,\ldots,x_n:=\sqrt{2x_{n-1}}$$

Show that $\;\{x_n\}\;$ is (1) bounded and (2) monotone ascending, and thus it has a limit. Now use arithmetic of limits.

Answer 3

Here's a non-rigorous way to find the answer. Call the final result $x$. Then $x^2 = 2 \cdot x$, and you can solve for $x$. This method can give the wrong answer for similar-looking problems: the iteration can converge to different values depending on the starting point, or it may not converge at all.