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Let $a_{1} = 3$ and let $a_{n+1} = \frac{3a_{n}+7}{a_{n}+3})$ if $n \geq 1$ is true Show that $a_{n}$ converges.

I am stuck in how to show tht it converges:

Do I just say that as $a_{n}\rightarrow a$ so then $\epsilon > 0$ then there is some $N >0$ such that if $n>N$ then $|a_{n} - a|<\epsilon$.

Is that allthere is too it?

Thanks for any help.

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    Ar you allowed to use something else than the definition of a convergent sequence? – 2017-02-16

1 Answers 1

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For $a_n>\sqrt7$, $$a_{n+1}=\frac{3a_n+7}{a_n+3}=3-\frac2{a_n+3}>\sqrt7$$

so the sequence is bounded below, while

$$a_{n+1}-a_n=\frac{7-a_n^2}{a_n+3}<0$$ and the sequence is decreasing.

The value $\sqrt7$ is nothing but the limit of $a_n$, found as the positive root of

$$a=\frac{3a+7}{a+3}.$$