Sequence Strictly Increasing

Definition

$\forall n\in \mathbb{N},a_n<a_{n+1}$

• This is proved by PMI

Example

Show $a_1=1,a_{n+1}=\sqrt{1+5a_n}$ is strictly increasing:

Soln

• Define $P(n):a_n<a_{n+1}$
• Proving base case:
  • $P(1)=1<\sqrt{6}=\sqrt{1+5(1)}=\sqrt{1+5a_n}$
• Proving induction step:
  • Suppose $\forall n\ge 1,P(n)$
  • Then, $a_n<a_{n+1}$ by defn of $P(n)$
  • Consider $a_{n+2}=\sqrt{1+5a_{n+1}}=\sqrt{1+5(\sqrt{1+5a_n)}}>\sqrt{1+5a_n}$ as $\sqrt{\cdot }$ is increasing by defn
  • Thus, $a_{n+1}<a_{n+2}$
  • Thus, $P(n)⟹P(n+1)$
• Thus, $\forall n\ge 1,a_n<a_{n+1}$