[College Calc 01-02. Limits of Sequences] #07

Let us take the base-2 logarithm on both sides of the recurrence relation:

$\log_2 a_{n+1} = \frac{1}{2} + \frac{1}{2}\log_2 a_n.$

Note that letting $b_n = \log_2 a_n$ we have a linear recurrence relation much like the one in this previous problem , with $b_0 = \log_2 a_0 = 0.$ Solving the relation -- refer to the solution of the linked problem if you don't know how -- we have

$b_n = -\left(\frac{1}{2}\right)^n + 1.$

Then, we have

$a_n = 2^{b_n} = 2^{-\left(\frac{1}{2}\right)^n + 1},$

so that the value we want is

$\lim_{n\to\infty}\left[\frac{1}{2^n}\prod_{k=0}^{n} a_k\right] = 2\lim_{n\to\infty} \prod_{k=0}^{n} 2^{-\left(\frac{1}{2}\right)^k} =2\cdot 2^{-\sum_{k=0}^{\infty}\left(\frac{1}{2}\right)^k} = 2\cdot 2^{-2} = \boxed{\frac{1}{2}}.$

Given that $a_0=1$ and $a_{n+1} = \sqrt{2a_n}$ , we have:

$\large \begin{aligned} a_0 & = 1 \\ a_1 & = 2^\frac 12 \\ a_2 & = 2^\frac 12 \cdot 2^\frac 14 = 2^{\frac 12 + \frac 14} \\ a_3 & = 2^{\frac 12 + \frac 14 + \frac 18} \\ \implies a_n & = 2^{\frac 12 \cdot \frac {1-\frac 1{2^n}}{1-\frac 12}} = 2^{1-\frac 1{2^n}} \\ \prod_{k=0}^n a_k & = 2^{\sum_{k=0}^n \left(1-\frac 1{2^k}\right)} = 2^{n+1 - 2 + \frac 1{2^n}} = 2^{n-1+\frac 1{2^n}} \\ \frac 1{2^n}\prod_{k=0}^n a_k & = 2^{\frac 1{2^n}-1} \\ \implies \lim_{n \to \infty} \frac 1{2^n}\prod_{k=0}^n a_k & = 2^{-1} = \boxed{\frac 12} \end{aligned}$

We see that $a_k=2^{1-\frac{1}{2^k}}$

So the given expression within the bracket equals

$\dfrac {1}{2^n}\times \dfrac {2^n}{2^{\frac 12+\frac{1}{2^2}+...+\frac{1}{2^n}}}$

$=2^{\frac 1n-1}$ , and the required limit is $2^{-1}=\boxed {\dfrac 12}$ .