Showing the convergence of the given recursively defined sequence Let $(a_1) = 1$ and $$(a_{n+1})= \frac{4+3a_n}{3+2a_n}$$
It is required to show that the following recursively defined sequence converges. I know one way to show this converges. Define $$f(x)= \frac{4+3x}{3+2x}$$ on the set of non-negative reals(which is closed hence complete). It is easy to show that $f'$ exists and $|f'(x)|<1$ hence it is a contraction and by the fixed point theorem has a unique fixed point. Is there any easy way to show that this converges using only the basic theorems on limits. All my attempts have failed so far. Any solutions will be highly appreciated.
 A: $$(a_{n+1})= \frac{4+3a_n}{3+2a_n}= 1+\frac{1+a_n}{3+2a_n}$$
the sequence is bounded increasing  so the limit exist and 
$$l= \frac{4+3l}{3+2l} \Rightarrow 3l+2l^2=4+3l \Rightarrow  l^2=2  \Rightarrow l=\pm \sqrt{2}$$
as the limit is positive so $l=\sqrt{2}$
A: This is a homographic sequence. The strategy is to express $a_{n+1}-a_n$ in function of $a_n-a_{n-1}$.
First rewrite $f(x)$ in canonical form (obtained by Euclidean division of $3x+4$ by $2x+3$:
$$\frac{4+3x}{3+2x}=\frac32-\frac1{2(3+2x)}.$$
We obtain
\begin{align}
a_{n+1}-a_n&=\biggl(\frac32-\frac1{2(3+2a_n)}\biggr)-\biggl(\frac32-\frac1{2(3+2a_{n-1})}\biggr)\\&=\frac1{2(3+2a_{n-1})}-\frac1{2(3+2a_n)} %\\[1ex]
=\frac{a_n-a_{n-1}}{(3+2a_{n-1})(3+2a_n)}.
\end{align}
As $a_1>0$, it is easy to see all $a_n$ are positive. from this relation, we draw two conclusions:


*

*$a_{n+1}-a_n$ and $a_n-a_{n-1}$ have the same sign for all $n$, hence the sequence is monotonic, and ac tually increasing since $a_2>a_1$.

*$0<a_{n+1}-a_n <\frac19(a_n-a_{n-1})$, so that $a_{n+1}-a_n <\frac1{9^{n-1}}(a_2-a_{1})$ and
$$a_n-a_1=\sum_{k=1}^{n-1}(a_{k+1}-a_k)<\sum_{k=1}^{n-1}\frac1{9^{k-1}}(a_2-a_1)<\frac98(a_2-a_1)=\frac9{20}$$
and finally $\; a_n=(a_n-a_1)+a_1<\frac{29}{20}$ is a bounded increasing sequence. Therefore it converges.


By continuity, the limit is a non-negative root of the equation
$$x=\frac{3x+4}{2x+3}\iff x^2=2.$$
