I want to know the limits and convergence/divergence of $a_{n+1} = (a_n+3)/(a_n+4)$ I`ve the following iteratively defined sequence:
$a_1=1$ and for $n\ge 1$,
$$a_{n+1} = \frac{a_{n}+3}{a_n+4}$$
I want to know the limits and if it is convergent or divergent, but I don't know how to handle two $a_n$'s in the formula.
If I have got one $a_n$ I would find it out by proving monotony and  boundedness. I think that I can use the same technique here but I got stuck. Thanks in advance.
 A: For convergence, just use Banach Fixed Point on $I=[0,2]$ and $f(x)=(x+3)/(x+4)$. Looking at the graph, you see that $f(I)\subset I$, and all that remains is to show that the map is contractive in a strong-enough sense. Take $x$ and $x'$ in $I$, and compute:
\begin{align}
f(x)-f(x')&=\frac{x+3}{x+4}-\frac{x'+3}{x'+4}\\
&=\frac{x-x'}{(x+4)(x'+4)}
\end{align}
Take absolute values, and get $|f(x)-f(x')|\le\frac1{16}|x-x'|$, plenty good enough. So there’s a fixed point in $[0,2]$, and @orlp’s computation of it as $\frac12(\sqrt{21}-3)$ gives the value.
A: After testing a couple $a_n$ by hand it appears to converge. So let's try to solve:
$$x = \frac{x+3}{x+4}$$
$$x(x + 4) = x+3$$
$$x^2 +3x - 3 = 0$$
$$x = \frac{1}{2}(\sqrt{21} - 3)$$
But if we want to be rigorous and prove its convergence:
$$a_n = \frac{p_n}{q_n}$$
$$\frac{p_{n+1}}{q_{n+1}} = \frac{\frac{p_n}{q_n}+3}{\frac{p_n}{q_n}+4} = \frac{p_n + 3q_n}{p_n + 4q_n}$$
$$P(x) = \sum_{n=1}^\infty p_nx^n \quad Q(x) = \sum_{n=1}^\infty q_nx^n$$
$$P(x) = x + x\sum_{n=1}^\infty p_{n+1}x^n = x + x\sum_{n=1}^\infty (p_n + 3q_n)x^n = x + xP(x) + 3xQ(x)$$
$$Q(x) = x + x\sum_{n=1}^\infty q_{n+1}x^n = x + x\sum_{n=1}^\infty (p_n + 4q_n)x^n = x + xP(x) + 4xQ(x)$$
$$Q(x) = \frac{x + xP(x)}{1 - 4x}$$
$$P(x) = x + xP(x) + 3x\cdot \frac{x + xP(x)}{1 - 4x} = \frac{x(1 - x)}{x^2 - 5x + 1}$$
After plugging $P(x)$ back in we find that $Q(x)$ is quite simple:
$$Q(x) = \frac{x}{x^2 -5x + 1} = \frac{1}{\beta - \alpha}\left(\frac{\alpha}{\alpha - x} - \frac{\beta}{\beta - x}\right)$$
Where $x^2 - 5x + 1 = (x - \alpha)(x - \beta)$ and thus $\alpha = \frac{1}{2}(5 + \sqrt{21})$ and $\beta = \frac{1}{2}(5 - \sqrt{21})$.
Knowing the formula for the inverse geometric series $\displaystyle \sum_{n=1}^\infty \alpha^{-n}x^n = \frac{a}{a-x}$ we get:
$$Q(x) =  \frac{1}{\beta - \alpha}\left(\sum_{n=1}^\infty (\alpha^{-n} - \beta^{-n})x^n\right)$$
$$q_n = \frac{\alpha^{-n} - \beta^{-n}}{\beta - \alpha}$$
But subtracting the original numerator from the denominator we find $q_{n+1} - p_{n+1} = q_n$. Thus
$$\dfrac{p_{n}}{q_{n}} = 1 - \dfrac{q_{n-1}}{q_{n}} = 1 - \frac{\alpha^{1-n} - \beta^{1-n}}{\alpha^{-n} - \beta^{-n}}$$
Finally, for large $n$ we find that $\alpha^{-n} \to 0$ as $\alpha > 1$, thus:
$$\lim_{n\to\infty} \dfrac{p_{n}}{q_{n}} = 1 - \frac{\beta^{1-n}}{\beta^{-n}} = 1 - \beta \approx 0.791288$$

As a fun fact we have:
$$p_n = \frac{(\alpha - 1)\alpha^{-n} - (\beta- 1)\beta^{-n}}{\alpha - \beta}$$
A: If we set $a_n=1-b_n$ we get $b_1=0$ and
$$ b_{n+1} = \frac{1}{5-b_n}.$$
Let $\lambda$ be the smallest positive root of $x=\frac{1}{5-x}$, i.e. $\lambda=\frac{5-\sqrt{21}}{2}=\frac{2}{5+\sqrt{21}}$ and $I=[0,\lambda]$.
The function $f:I\to I$ defined by $ f(x)=\frac{1}{5-x}$ is increasing and Lipschitz-continuous.
It gives a contraction of $I$ since $|f'|\leq\frac{1}{16}$, hence by the Banach fixed point theorem the sequence $\{b_n\}_{n\geq 0}$ converges to the only fixed point of $f$ in $I$, i.e. $\lambda$. As a consequence,
$$ \lim_{n\to +\infty}a_n = 1-\lambda = \frac{\sqrt{21}-3}{2}.$$
