What is the connection with $g=1+\frac{1}{g}$ and $G_n= 1+\frac{1}{G_{n-1}}$? I'm reading George E. Martin's: The art of enumerative combinatorics. In here:



I don't understand "in the limit". I know that $g=1+\frac{1}{g}$ has a connection with $G_n =1+\frac{1}{G_{n-1}}$ because they are similar, but it's not very clear how to reach the former using the latter.

For each $n$, the real numbers $G_n$ and $1+1/G_{n-1}$ are the same. So, if one of them converge, so does the other, and they converge to the same real number. 
Well, if you define $g=\lim G_n$, then $g$ is also the limit of $G_{n-1}$. Since we know that: $$\lim G_n = \lim \{1+1/G_{n-1}\}$$ $$\lim G_n = g$$ $$\lim\{1+1/G_{n-1}\}=1+1/g$$
we discover that: $$g=1+1/g$$
This whole argument is usually omitted, and we just say "let $n\to\infty$ in the equation $G_n=1+1/G_{n-1}$".
