# Help with proof by induction of inequality [duplicate]

This question already has an answer here:

I am studying for an exam and going through various earlier tutorial sheet questions. For the question below, I have tried and just can't figure out how to prove that $x$n < $3$ by mathematical induction. Does anyone know how to prove this?

The question is:

Let $x$1 = 1, and for each $n \in \mathbb{N}$ let $x$n+1 = $\frac{2}{3}x$n + $1$. Then $x$n < $3$ for all $n \in \mathbb{N}$.

## marked as duplicate by choco_addicted, pjs36, Community♦Mar 26 '16 at 6:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• If $x_n<3$, then what can you say about $\frac{2}{3}x_n+1$? – carmichael561 Mar 26 '16 at 5:43

## 1 Answer

Hint : For the inductive step, you have

\begin{align}x_{n + 1} &= \frac{2}{3}x_n + 1\\&<\frac{2}{3}(3) + 1\\&= 3\end{align}

• A bit more than a hint, I'd say. – carmichael561 Mar 26 '16 at 5:44
• I tried - It probably can't get any more brief than this ;) – Yiyuan Lee Mar 26 '16 at 5:45