Proving convergence of a sequence $a_{n+1} = 3 - 2/a_n$ and finding the limit. 
Let $(a_n)$ be the sequence defined by: $$a_1=\frac{3}{2}\qquad a_{n+1}=3-\frac{2}{a_n}\quad\text{for all }n.$$

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*Prove that the sequence is convergent.

*Calculate the limit of $(a_{n+1})$.


I do 1. by induction. Since $(a_1)$ is convergent, assume $(a_n)$ is convergent. $(a_{n+1})$ is convergent, proved from limit arithmetic.
Re 2., I think that the limit is $0$, from the rule:
$$\left|\frac{a_{n+1}}{a_n}\right|<1 \implies \lim_{n\to\infty}{a_n}=0$$
Am I right? What are the ways of doing this?
Note: I haven't studied Taylor Series yet.
 A: Since $a_1 < a_2$, assume that $a_n < a_{n+1}$ then $$\frac{2}{a_n} > \frac{2}{a_{n+1}} \iff - \frac{2}{a_n} < -\frac{2}{a_{n+1}} \iff 3 - \frac{2}{a_n} < 3- \frac{2}{a_{n+1}}$$ which is equivalent to $a_{n+1} < a_{n+2}$ hence the sequence is monotone increasing by induction. We need only show that it is bounded above by $2$ now. To do so, we induct. $a_1 \leq 2$, assume that $a_n \leq 2$ then $$-\frac{2}{a_n} \leq -1 \iff 3 - \frac{2}{a_n} \leq 2$$ which is equivalent to $a_{n+1} \leq 2$. Hence the sequence is bounded above by $2$.
Since the sequence is both bounded and monotone, then by the monotone convergence theorem the sequence converges to some limit $\ell$. We also know that $\lim a_n = \lim a_{n+1} = \ell$
Then we use limit arithmetic to get that the sequence converges to $2$. If the sequence converges to $\ell$ then $$\ell = 3 - \frac{2}{\ell} \implies \ell = 2$$ We also have $\ell = 1$, but this cannot be true since $a_1 > 1$ and the sequence is monotone increasing. 
A: By using $a_{1} = \frac{3}{2}$ and the difference equation
\begin{align}
a_{n+1} = 3 - \frac{2}{a_{n}} \hspace{5mm} n \geq 1
\end{align}
then by writting out the first few terms it can be seen that
\begin{align}
a_{n+1} = 1 + \frac{2^{n}}{2^{n}+1}.
\end{align}
Taking the limit as $n \to \infty$ then:
\begin{align}
\lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \left\{ 1 + \frac{1}{1 + \frac{1}{2^{n}}} \right\} = 2
\end{align}
A: The first thing that we observe is that $$a_n = \frac {2^n + 1} {2^{n - 1} + 1} = 2 - \frac {1} {2^{n - 1} + 1}.$$ This means that $a_n$ converges and to $2$.
