Find a formula for $a_n$ that satisfies $a_0 = 1$ and $a_{n} = 2 \sqrt{a_{n-1}}$ if $ n > 0$

Question

$a_{n} = 1 \ \ when \ \ n = 0 , and \ \ a_{n} = 2 \sqrt{a_{n-1}} \ \ when \ \ n > 0$

I'm a complete beginner at solving something like this, and I'm not sure where to start. The two methods I know are iterative expansion which I have trouble understanding how it works. And I know how to use induction, but I'm not sure how to apply it here.

Now just by doing a few terms by hand (hopefully correctly) here's what I got,

$ 1, 2, 2\sqrt{2}, 2\sqrt{2\sqrt{2}}$ and I'm pretty sure it just has n amount of $2\sqrt{2}$ under the sqrt sign.

I'm not sure how to progress with these types of problems in general. Thanks.

Answer 1

$$a_n=2\sqrt{2\sqrt{2\sqrt{...}}}$$ $$=2^{(1+\frac12+\frac14+...+\frac1{2^{n-1}})}$$ $$=2^{\bigg(\frac{1-\frac1{2^n}}{1-\frac12}\bigg)}$$ $$=2^{2(1-\frac1{2^{n}})}$$ $$=4^{(1-\frac1{2^{n}})}$$

Answer 2

You've found the sequence begins $2^0,\,2^1,\,2^{3/2},\,2^{7/4}$. You may prove by induction that $a_n=2^{2-2^{1-n}}$. (It helps to first define $b_n:=\log_2 a_n$ so $b_0=0,\,b_{n+1}=\frac{b_n}{2}+1$, implying $b_n$ differs from the fixed point $2$ of $x\mapsto\frac{x}{2}+1$ by a geometric sequence.)

Answer 3

Divide it by 4 and we get $${a_n\over 4} =\sqrt{a_{n-1}\over 4}$$

Let $b_n = a_n/4$ then we have $$b_n = \sqrt{b_{n-1}}$$ $$=\sqrt[2^2]{b_{n-2}}$$

$$=\sqrt[2^3]{b_{n-3}}$$ $$=\sqrt[2^4]{b_{n-4}}$$ $$\vdots $$ $$=\sqrt[2^n]{b_{0}}$$

Since $b_0= a_0/4 = 1/4$ we have $$a_n = 4\cdot \sqrt[2^n]{{1\over 4}}$$