$a+\dfrac {a+\dfrac {a+\dfrac {a+\dfrac {:} {b}} {b}} {b}} {b}=?$
I've tried letting $\quad a+\dfrac {a+\dfrac {a+\dfrac {:} {b}} {b}} {b}=K$
Which makes the equation:
$a+\dfrac {K} {b}=K$ $\quad$ and $\quad$ $a=\dfrac {bK-K} {b}$ $\quad$ and it'll be $\quad\dfrac {ab} {b-1}=K$
How we can say that is true?$\;$Where does that logic come from?
$a+\dfrac {a+\dfrac {a+\dfrac {a+\dfrac {:} {b}} {b}} {b}} {b}=\dfrac {ab} {b-1}$
And other examples:
$\sqrt [n] {a\sqrt [n] {a\sqrt [n] {a\sqrt [n] {a\ldots }}}}=\sqrt [n-1] {a}\tag{1}$
$\sqrt [n] {a:\sqrt [n] {a:\sqrt [n] {a:\sqrt [n] {a:\ldots }}}}=\sqrt [n+1] {a}\tag{2}$
$\sqrt [] {a+\sqrt [] {a+\sqrt [] {a+\sqrt [] {a+\ldots }}}}=\dfrac {1+\sqrt {1+4a}} {2}\tag 3$
$\sqrt [] {a(a+1)+\sqrt [] {a(a+1)+\sqrt [] {a(a+1)+\sqrt [] {a(a+1)+\ldots }}}}=a+1 \tag4$
$$\boxed{\boxed{\text{HOW ??}}}$$
The way this is handled in calculus courses is that you have to show that the sequence is, say monotone increasing and bounded from above. Then applying the least upper bound property of the real numbers you can conclude that the sequence converges. Once you know it converges you can apply the argument you gave. However, checking boundedness and monotonicity can be the trickier part of the proof.