Proving $a_n>2$ where $a_k=\sqrt{2a_{k-1}}$

Question

Need to prove that $a_n > 2$ in the sequence: $a_k = \sqrt{2a_{k-1}}$ for $k>1$.

$a_1 = 3$.

Not sure how to use the inductive hypothesis: $P(k) = \sqrt{2a_{k-1}}$ to find $P(k+1)$.

Answer 1

You first need to show your base case. For $k = 2$, we have $$a_2 = \sqrt{2 a_1} = \sqrt{2 \cdot 3} = (\sqrt{2})(\sqrt{3}) > (\sqrt{2})(\sqrt{2}) = 2$$

Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.

We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have, \begin{align} a_{k+1} &= \sqrt{2 a_k}\\ &= (\sqrt{2})(\sqrt{a_k})\\ \end{align} But by the inductive hypothesis, we have that $a_k > 2 \Rightarrow \sqrt{a_k} > \sqrt{2} \Rightarrow (\sqrt{2})(\sqrt{a_k}) > (\sqrt{2})(\sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.

Hence by the principle of mathematical induction, we are done :)