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The sequence is defined by:

$a(1)=1, a(n+1) = \frac{3(1+a(n))}{3+a(n)}$

How do I show that the sequence is monotonic and bounded from above?

My approach to show its monotone was:

$a(n+1) - a(n) \geq 0$

$ \frac{3(1+a(n))}{3+a(n)} - a(n) = \frac{3+3a(n)}{3+a(n)} - \frac{3a(n)+ a(n)^2}{3+a(n)} = \frac{3(1+a(n)) - a(n)(3+a(n))}{3+a(n)} = \frac{3-a(n)^2}{3+a(n)} \geq ? \geq 0$

I know the sequence is bounded from above by $\sqrt{3}$ but I'm totally lost at how to show this.

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2 Answers 2

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Use induction to show that $a(n) <\sqrt 3$ for all $n$. Then use the fact that $\frac {3(1+x)} {3+x} >x$ for $0<x <\sqrt 3$ to show that $a(n+1) >a(n)$ for all $n$. You can also check that $a(n) <2$ for all $n$ so $(a(n))$ is convergent.

I will let you show that the limit is $1$.

[It is useful to start with the observation that $\frac {3(1+x)} {3+x}$ is an increasing function on $(0,\infty)$].

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Notice that $$\dfrac{a_{n+1}-\sqrt{3}}{a_{n+1}+\sqrt{3}}=\dfrac{a_{n}-\sqrt{3}}{a_{n}+\sqrt{3}}\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$$. We can calculate the accurate value of $a_n$.

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