Let be $a,b$ positive real numbers; $a+\frac {b} {a+\frac {b} {a+\frac {b} {\ddots }}}$ how can we prove whether this has a limit or not? Let, given example is like this

$$3+\dfrac {4} {3+\dfrac {4} {3+\dfrac {4} {\ddots }}}=?$$

I wonder if it has a limit or not, but I have a idea, If we show that this sequence monotonous and has a bound, we can say that this has a limit.
Therefore,firstly, we need to write sequence form;

$$3+\dfrac4{x_n}=x_{n+1}$$

If we assume that this sequence has a limit, we can apply convergence of subsequences theorem, with respect to this theorem we can do the following conclusion.Let assume $x$ equals to limit of the sequence>$$3+\dfrac4{x}=x$$
$x_1=4$ and $x_2=-1$, limit should be positive, so $x=4$.That is;
$$3+\dfrac {4} {3+\dfrac {4} {3+\dfrac {4} {\ddots }}}=4$$

My intuition says me that the sequence is increasing and has a upperbound but I couldn't show, so above proof is not valid for now.

How we can that this is a monotonous and has a bound?
 A: The sequence is not monotonic. In fact successive terms must alternate between being greater than or less than $4$, so the sequence alternates going up and down.
What is true is that the even terms are increasing and the odd terms decreasing, or vice versa. Suppose $0<x_n<4$. Then $x_{n+2}=3+\frac{4x_n}{3x_n+4}$. We claim that $x_n<x_{n+2}<4$. The upper bound is easy: $x_n<4$ so $4x_n<3x_n+4$. The lower bound is equivalent to $\frac{4x_n}{3x_n+4}>x-3$, and since $3x_n+4>0$ this is equivalent to $(x_n-3)(3x_n+4)<4x_n$, i.e. $3(x_n+1)(x_n-4)<0$, which is true since $x_n<4$.
This shows that if $0<x_0<4$ then the even terms will tend to a limit $x$. This easily implies that the odd terms tend to $3+4/x$, and so the whole sequence will converge if $x=4$. To prove this, try to adapt the above to show that if the limit of the even terms is less than $4$ then their differences don't tend to $0$, which gives a contradiction.
A: One way to prove convergence is to start with the assumed limit and consider the distance to it.
As you've seen the equation for the assumed limits is a quadratic so we could instead start in the other end and write the recurrence formula:
$$x_{n+1} = f(x) = (p+q) - pq/x_n$$
where $p$ and $q$ are the assumed limits. Now as the iteration is a solution method for the fix-value problem it will converge if the secant to the fixpoint has absolute slope of always less than $1$ (since that would mean that the distance to the solution will decrease in every step).
Now the slope to a fixpoint is:
$${f(x) - f(p)\over x-p} = {p+q-pq/x-p\over x-p} = {q(x - p)\over x(x-p)} = {q\over x}$$
So if always $|q/x_n|<1-\epsilon$ the series will converge to $p$. And if always $|q/x_n|>1+\epsilon$ it will diverge (and do so strictly).
This works on your example since you will have with $p=4$ and $q=-1$ that if $|x-p|\le 1$ that $x \ge 3$ which means that $|q/x| = 1/|x| \le 1/3$. As long as $|x-p|\le 1$ it will approach $p$ strictly and therefore it the $|x-p|\le 1$ will remain valid and it will aproach $p$ strictly indefinitely.
However the sequence is not monotone (as you can easily see by testing). What I mean by strict approaching is that $|x_n-L|$ is strictly decreasing, strict convergence to $L$ is convergence to $L$ that strictly approaches $L$.
A: Another way to prove convergence to $l=\frac{a}2+\sqrt{\frac{a^2}4+b}$ (the general case) is to show that the absolute difference each successive member of the sequence has to $l$ decreases each time, which works in some cases:
Start with any $x_1 > 0$. You get
$$ \left| x_{n+1}-l\right| = \left| a+\frac{b}{x_n} - l\right| = \left|\frac{2ax_n+2b-(a+\sqrt{a^2+4b})x_n}{2x_n}\right| = \left|\frac{2b+(a-\sqrt{a^2+4b})x_n}{2x_n}\right| = \left|\frac{\frac{2b}{(a-\sqrt{a^2+4b})} +x_n}{\frac2{(a-\sqrt{a^2+4b})}x_n}\right| = \left|\frac{-\frac{(a+\sqrt{a^2+4b})}2 +x_n}{\frac2{(a-\sqrt{a^2+4b})}x_n}\right| = \left| \frac{-l+x_n}{\frac2{(a-\sqrt{a^2+4b})}x_n}\right| =  \frac{\left|x_n-l\right|}{\frac2{\sqrt{a^2+4b}-a}x_n}$$.
If the denominator in the last expression is always known to be at least some fixed $q > 1$, this proves convergence. Since $x_n>a$, this is satisfied if
$$\frac2{\sqrt{a^2+4b}-a}a >1,$$
which simplifies to $b < 2a^2$, which is true in your example $a=3,b=4$. 
In reality, it looks like the iteration will always converge, even if $b \ge 2a^2$. What happens is that it takes a while for the iteration to get into the mode where each iteration is better then the previous, though I can't strictly prove that.
