$a_{n + 1} = 5 - \frac{6}{a_n + 2}$ with $a_1 = 1$ . Prove by induction that $a_n < 4$ for $n \geq 1$.

Question

So I got this induction proof question but I can't seem to make a logical statement in one part of it:

The question is , $a_{n + 1} = 5 - \frac{6}{a_n + 2}$ with $a_1 = 1$ . Prove by induction that $a_n < 4$ for $n \geq 1$

I reached up to the proof where I need to prove $a_{k+1} <4$

Proof

$a_k <4 \implies a_k + 2<6 $

The next step I want to put is:

$\frac{6}{ a_k +2} >1$

However I can only justify this statement if $a_k > -2$ but I can't seem to prove that or find any info in the question to suggest that it.

Can anyone help me with the proof or my theory?

Answer 1

Assuming you have shown the base case, suppose $0< a_k < 4$ for some $k \in \mathbb{N}$. Then we have that $0< a_k + 2 < 6$, which implies \begin{align*} \frac{6}{a_k + 2} > 1 \end{align*} (divide both sides of the inequality by $a_k + 2$). Also notice that $\frac{6}{a_k + 2} < 3$ since we have assumed $a_k > 0$ and hence $a_k + 2 > 2$.
\begin{align*} 0 < a_{k+1} = 5 - \frac{6}{a_k + 2} < 4 \end{align*} as desired.

Answer 2

Hint:   write it as $\;a_{n + 1} -4 = \dfrac{a_n-4}{a_n + 2}\,$, then (prove and) use that $\,a_n+2 \gt 0\,$.