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?

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.
