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.
Best Answer
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.