Determine convergence of sequence $a_{n+1}=\frac{7+3a_n}{3+a_n}$ I want to find out if the following sequence is convergent:
$$
a_1 = 3,\quad
a_{n+1} = \frac{7 + 3 a_n}{3 + a_n}, \quad (n ≥ 1)
$$
I am new at university and it would be great if someone has got a tip for me how to deal with this. THX a lot.
 A: You can rewrite $a_{n+1} = 3 - \frac{2}{3+a_n}$ so that, if you know $a_{n+1} < a_{n}$ then $(3+a_{n+1})^{-1} > (3+a_n)^{-1}$ and so $$3 - \frac{2}{3 + a_{n+1}} < 3 - \frac{2}{3+a_n} \iff a_{n+2} < a_{n+1}$$ so you can conlude it is  decreasing by induction (just check the base case). 
It's also bounded below by $0$. If $a_n > 0$ then $3+a_n >3$ so $3-\frac{2}{3+a_n}> 3-\frac{2}{3} > 0$ and you're done by induction again. 
So $a_n$ forms a decreasing sequence that is bounded below and hence converges. In fact it will converge to the unique positive $\ell$ that satisfies $\ell = (7+3\ell)(3+\ell)^{-1}$, that is $\ell = \sqrt{7}$.
A: Let $$f(x)=\frac{7+3x}{3+x}.$$ Show by induction that $a_n>0$ for all $n$. Now, if it converge it converge to a fix point of $f$, in particular $\sqrt 7$. By the way,
$$|a_n-\sqrt 7|=|f(a_{n-1})-f(\sqrt 7)|=\frac{2|a_{n-1}-\sqrt 7|}{(3+\sqrt 7)|a_{n-1}+3|}\leq \frac{1}{6}|a_{n-1}-\sqrt 7|,$$
where I used the fact that $a_{n}+3\geq 3$ and $3+\sqrt 7\geq 4$ for the last majoration. At the end,
$$|a_{n}-\sqrt 7|\leq \frac{1}{6^{n-1}}|3-\sqrt 7|\underset{n\to \infty }{\longrightarrow } 0,$$
and the claim follow.
