Proving by induction that for all n in set N an<3 is true

Question

Data : $a_0 = 2$. and for all n that in $N$ set $a_{n+1} = \sqrt{3*a_n}$.

prove that for every $n$ in $N$ $a_n<3$.

Now I know I need to use induction.

I did the first step and said that if $a_0=2$ and $2<3$ is really true, thats the base of my induction.

but I know I also need to use the series definition to prove the next step of the induction but im stuck there.

Answer 1

Let's assume the hypothesis is true for some $a_n$, i.e. $a_n < 3$. We need to verify it is also true for $a_{n+1}$. But $$ a_{n+1} = \sqrt{3 a_n} = \sqrt{3} \cdot \sqrt{a_n}. $$ Since $a_n < 3$ you have $\sqrt{a_n} < \sqrt{3}$...

Answer 2

Here is the induction step spelled out.

Assuming $0<a_n<3$, show that $0<a_{n+1}=\sqrt{3a_n}<3$

(I added the $0<{}$ just to make absolutely certain that the square root is allowed.)