Help with mathematical induction involving a sequence $ (A_{n})_{n \in \mathbb{N}} $.

Question

I need help on the following problem.

Problem. Let $ A_{1} \stackrel{\text{df}}{=} 2 $, and define $ A_{n + 1} \stackrel{\text{df}}{=} \dfrac{2 A_{n} + 1}{5} $ for all $ n \in \mathbb{N} $. Use mathematical induction to prove that $ A_{n} > \dfrac{1}{3} $ for all $ n \in \mathbb{N} $.

Answer 1

For $n = 1$, $A_n = 1 > \frac{1}{3}$.

Assuming $A_k > \frac{1}{3}$, we have $2 A_k + 1 > \frac{5}{3}$, and so $A_{k+1} = \frac{2 A_k + 1}{5} > \frac{1}{3}$.

Answer 2

Hint: By the inductive hypothesis

$$ \dfrac{2A_n+1}{5} > \dfrac{2(\frac{1}{3})+1}{5}$$