Limit $ \sqrt{2\sqrt{2\sqrt{2 \cdots}}}$

Question

Find the limit of the sequence $$\left\{\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots\right\}$$

Another way to write this sequence is $$\left\{2^{\frac{1}{2}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}}2^{\frac{1}{8}},\hspace{5 pt} \dots,\hspace{5 pt}2^{\frac{1}{2^{n+1} - 2}}\right\}$$

So basically we have to find $\lim_{n\to\infty}2^{\frac{1}{2^{n+1} - 2}}$. This equates to $$\lim_{n\to\infty}2^{\frac{1}{2^{\infty+1} - 2}} \Longrightarrow 2^{\frac{1}{2^{\infty}}} \Longrightarrow 1$$

Is this correct? P.S. Finding that $S(n)$ was a pain!

Answer 1

Hints:

$$x=\sqrt{2\sqrt{2\sqrt2\ldots}}\implies x=\sqrt{2x}\implies x^2=2x\implies\ldots$$

Answer 2

$$\frac 1 2 + \frac 1 4 + \frac 1 8 + \dots = 1 $$

Answer 3

Hint: take a logarithm of every element in your sequence.

Answer 4

An alternative proof (of sorts):

If the sequence has a limit, then it will be a fixed point of $f(x)=\sqrt{2x}$, and hence a solution to $x=\sqrt{2x}$. The solutions are clearly $x=0, 2$, and you should be able to show that 0 is out because all of the terms in the sequence are $\gt 1$ and hence the limit is $\geq 1$.