(You can find out more about continued fractions here.) Now back to our sequence. The ratio of successive terms is $$\frac{a_i}{a_{i-1}}.$$ By equation (1) this is equal to $$\frac{a_i}{a_{i-1}}=\frac{na_{i-1}+a_{i-2}}{a_{i-1}}=n+\frac{a_{i-2}}{a_{i-1}}.$$ Now $$\frac{a_i}{a_{i-1}}=n+\frac{a_{i-2}}{a_{i-1}}=n+\frac{1}{\frac{a_{i-1}}{a_{i-2}}}.$$ Applying equation (1) to the term $a_{i-1}$ we get $$\frac{a_i}{a_{i-1}}=n+\frac{1}{\frac{na_{i-2}+a_{i-3}}{a_{i-2}}}=n+\frac{1}{n+\frac{a_{i-3}}{a_{i-2}}}.$$ You can see the picture: continuing to make substitutions according to equation (1), we end up with a finite continued fraction expression for $a_{i}/a_{i-1}$ of the form $$a_{i}/a_{i-1}=n+\frac{1}{n+\frac{1}{n+\frac{1}{...+\frac{a_1}{a_2}}}}.$$ Letting $i$ tend to infinity we can then prove that the ratio of successive terms in the sequence converges to infinite \emph{continued fraction} $$n+\frac{1}{n+\frac{1}{n+\frac{1}{n+...}}}. $$ Which, as we have sketch-proved above, is equal to $\lambda_n.$
Suppose we have an infinite sequence $$a_1,a_2,a_3,a_4,...$$ and that for all $i>2$ we have
\begin{equation}a_i=na_{i-1}+a_{i-2}, \end{equation} where $n>1$ is a positive integer.
In the main article we claimed that in this case the ratio of successive terms of the sequence converges to the metallic mean
$\lambda_n$. We'll now give you a justification of this claim. Recall that \begin{equation}\lambda_n=n+\frac{1}{\lambda_n}.\end{equation}
Substituting the expression given by this equation for the $\lambda_n$ in the denominator on the right hand side of the same expression gives
$$\lambda_n=n+\frac{1}{n+\frac{1}{\lambda_n}}. $$ Another substitution gives
$$\lambda_n=n+\frac{1}{n+\frac{1}{n+\frac{1}{\lambda_n}}}. $$
Continuing on in this vein, it is not too hard to prove that $\lambda_n$ is equal to the infinite \emph{continued fraction}
$$n+\frac{1}{n+\frac{1}{n+\frac{1}{n+...}}}. $$