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$a+\dfrac {a+\dfrac {a+\dfrac {a+\dfrac {:} {b}} {b}} {b}} {b}=?$

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:

$\sqrt [n] {a\sqrt [n] {a\sqrt [n] {a\sqrt [n] {a\ldots }}}}=\sqrt [n-1] {a}\tag{1}$

$\sqrt [n] {a:\sqrt [n] {a:\sqrt [n] {a:\sqrt [n] {a:\ldots }}}}=\sqrt [n+1] {a}\tag{2}$

$\sqrt [] {a+\sqrt [] {a+\sqrt [] {a+\sqrt [] {a+\ldots }}}}=\dfrac {1+\sqrt {1+4a}} {2}\tag 3$

$\sqrt [] {a(a+1)+\sqrt [] {a(a+1)+\sqrt [] {a(a+1)+\sqrt [] {a(a+1)+\ldots }}}}=a+1 \tag4$

$$\boxed{\boxed{\text{HOW ??}}}$$

Colbi
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Micheal Brain Hurts
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    Before you can start wondering what the value of the first expression is, you need to define what it means. A priori, things with dots in them do not necessarily mean anything until equipped with a definition. Once you decide on a definition (probably as a limit of a certain sequence), your reasoning shows at most that if there is a value, then that value with be $ab/(b-1)$; but it will require a separate proof to show that this is indeed the value. – hmakholm left over Monica Jun 05 '16 at 14:49

1 Answers1

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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.

Mikhail Katz
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