<s>[INST] Show that {$a_n$} is convergent and find sup{$a_n| n \in Z_+ $} $a_1 = 1$ and $ a_{n+1} = \frac{4+3a_n}{3+2a_n} ; \forall n \in Z_+$
Show that {$a_n$} is convergent, find its limit and find sup{$a_n| n \in Z_+ $} if exists.
I found the limit as follows -
Let$$\lim_{n \to \infty} a_n = L$$
$$\lim_{n +1\to \infty} a_n+1 = L$$
$$L = \frac{4+3L}{3+2L}$$ which gives$ L= \sqrt{2}$
$$\therefore \lim_{n \to \infty} a_n = \sqrt{2}$$
${a_n} $ is convergent ( $\sqrt{2} \in \mathbb R$)
Is there a fault in this? Can I straightaway say that $\sup\{a_n| n \in Z_+ \} =\sqrt{2} $  ?
 [/INST] $a_{n+1} - a_n = \dfrac{4+3a_n}{3+2a_n} - a_n = \dfrac{4-2a_n^2}{3+2a_n}$. So consider $f(x) = \dfrac{4+3x}{3+2x}$, we have: $f'(x) = \dfrac{1}{(3+2x)^2} > 0$, thus $f$ increases strictly. Next assume by induction that : $0 < a_n < \sqrt{2}$, then: $a_{n+1} = f(a_n) < f(\sqrt{2}) = \dfrac{4+3\sqrt{2}}{3+2\sqrt{2}} = \sqrt{2}$. Thus by induction $a_n < \sqrt{2}$, $\forall n$. Also this implies: $a_{n+1} - a_n = \dfrac{4-2a_n^2}{3+2a_n} > 0$. Thus: $0 < a_n < a_{n+1} < \sqrt{2}$, $\forall n$. Thus: $a_n$ increases and is bounded hence convergent to $L$. You solved $L = \sqrt{2}$, and this should imply that $\sqrt{2} = L = \sup\{a_n:n\in \mathbb{N}\}$
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