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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 $n \in \mathbb{N}$.

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

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  • $\begingroup$ If $x_n<3$, then what can you say about $\frac{2}{3}x_n+1$? $\endgroup$ – carmichael561 Mar 26 '16 at 5:43
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Hint : For the inductive step, you have

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

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

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