# Induction proof inequality

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?

## 2 Answers

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

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.

• I appreciate your help but can you justify how the implication ( <1) is true? I can't seem to wrap my head around it – user122343 Aug 2 '18 at 6:04
• If ak was 2 for example , the inequality won't hold – user122343 Aug 2 '18 at 6:09
• Sorry I made a mistake. It should be 6/(a_k + 2) >1. – matt stokes Aug 2 '18 at 6:11
• If ak was -3 wouldn't the statement not hold also . I'm a bit confused on this since I don't know for sure if ak itself ( the terms of the sequence ) is greater than 0 . I assume it is from plugging in values but I feel as though I can't just write that worhout proving for all – user122343 Aug 2 '18 at 6:14
• Notice a_1 = 1 and the statement holds (this is the base case). Now for the induction part of the proof, if we assume a_n < 4 for arbitrary n, we have just shown that a_{n+1}<4. This implies that a_n can never be -3. – matt stokes Aug 2 '18 at 6:19