Showing the sequence $\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$ tends to $2$ using the Epsilon-Neighbourhood definition I am given the sequence
$$a_{n+1} = \sqrt{2 a_n}, \quad a_1 = \sqrt{2}.$$
That is, the sequence
$$ \sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$$
I am aware of using the recursive method to find the limit: setting $x = \sqrt{2x}$ and getting $x = 2$, rejecting $x = 0$ due to the sequence being monotone and positive.
I now want to show that the limit of this sequence is $2$ using the epsilon-neighbourhood definition for a convergent series:

Let $(a_n)$ be a sequence that converges to a real number $a$. Then, for every number $\epsilon > 0$, there exists a number $N \in \mathbb{N}$ such that for all $n \geq N$, $|a_n - a| < \epsilon$.

I take the logarithm on both sides and arrive at
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \ln 2 - \ln 2 \right|  < \epsilon. $$
Using triangle inequality, $|x - y| \geq |x| - |y|$, I arrive at
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \right| - |\ln 2| < \epsilon. $$
Add $\ln 2$ to both sides
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \right| < \epsilon + \ln 2. $$
The left hand side can be rewritten as a sum of $n$ terms in a geometric series with first term $\frac{1}{2}$ and common ratio $\frac{1}{2}$,
$$ \frac{\frac{1}{2} \left( 1 - \left( \frac{1}{2} \right)^n \right)}{1 - \frac{1}{2}} = 1 - \frac{1}{2^n} < \epsilon + \ln 2. $$
I am unsure of how to proceed beyond this point.
 A: Let's make an observation: for each natural number $n$, your sequence can be given explicitly by $$a_n = 2^{\sum_{i=1}^n (\frac{1}{2})^i}$$
For example, $a_2 = \sqrt{2\sqrt{2}} = (2(2^{1/2}))^{1/2} = 2^{1/2} \cdot 2^{1/4} = 2^{1/2 + 1/4}$
Furthermore, we know that $${\sum_{i=1}^n \left(\frac{1}{2}\right)^i} \to 1 \ \ \text{as} 
 \ \ n \to \infty$$
We also know that the function $f(x) = 2^x$ is a continuous function. In other words, if we pick $\epsilon > 0$, there is $\delta > 0$ such that $|x-1| < \delta$ implies $|f(x) - f(1)|< \epsilon$.
Thus, because $$\sum_{i=1}^n \left(\frac{1}{2}\right)^i \to 1$$there is $M > 0$ so that $n > M$ implies $$\left|{\sum_{i=1}^n \left(\frac{1}{2}\right)^i -1}\right| < \delta$$
Hence, letting $x = \sum_{i=1}^n \left(\frac{1}{2}\right)^i$ in the paragraph where I mentioned continuity, we obtain $$\left|f\left(\sum_{i=1}^n \left(\frac{1}{2}\right)^i\right) - 2\right| < \epsilon$$
This is precisely $|a_n -2 | < \epsilon$
A: Hint:
For $n>1$, we can write
$$\frac{a_n}2=2^{-2^{-n}}>2^{-\frac1n}>1-\frac1n$$ because $2^n>n$ and because by the binomial theorem
$$\left(1-\frac1n\right)^{-n}>\left(1+\frac1n\right)^n=1+\frac nn+\frac{n(n-1)}{2n^2}+\cdots+1>2.$$
Now it suffices to prove convergence of $1-\dfrac1n$ to $1$, which is easy, and will allow to write
$$n>N\implies\left|2-\frac2n\right|<\epsilon\implies\left|2^{1-2^{-n}}-2\right|<\epsilon.$$
The figure illustrates the first inequalities.

