*continued fraction*$$n+\frac{1}{n+\frac{1}{n+\frac{1}{n+...}}}. $$

(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 *continued fraction*
$$n+\frac{1}{n+\frac{1}{n+\frac{1}{n+...}}}. $$
Which, as we have sketch-proved above, is equal to $\lambda_n.$