closed form $f_n=\sqrt{2f_{n-1}}$ ?

Question

I am trying to write up a proof for the convergence of this recursive function. I was wondering if there exists a closed form.

Given first term in sequence is $\sqrt{2}$ and second is $\sqrt{2\sqrt{2}}$, and so on.

Answer 1

I get that

$$f_n = 2^{1/2 + 1/2^2 + 1/2^3 + \cdots+1/2^n} $$

or

$$\log_2{f_n} = \frac12 \frac{1-2^{-n}}{1/2} = 1-2^{-n}$$

i.e.,

$$f_n = 2^{1-2^{-n}}$$

Answer 2

Can you write $f_n = 2^{p(n)}$, where $p$ is some function of $n$?

Answer 3

If the first term is $\sqrt(2)$, it is easily shown that the $n$-th term is given by $f(n) = 2^\frac{2^n-1}{2^n}$

It is clear the exponent will approach 1 as n approaches infinity, and therefore $f(n)$ will aproach 2 as n approaches infinity.

Answer 4

I think you mean f_n =sqrt(2*f_n-1), given you say "successive terms" in the question.

Well you can break out

f_n = sqrt(2*f_n-1) 
f_n = 2^(1/2) * (f_n-1)^(1/2)

And close to

f_n = 2^(1/2n) * (f_0)^(1/2n)

And with your first term sqrt(2) this is

f_n = 2^(1/2n) * 2^(1+1/4n) 
f_n = = 2^(1+3/4n)

Then you can take a limit from that