Express $a_n$ in $n$: Recursive Formula

[Albert1]

$a_0=1, a_n=\dfrac {a_{n-1}}{1+(n-1)\times a_{n-1}}$
for all $n\geq 1$
express $a_n$ in $n$

 

[MarkFL]

My solution:

Spoiler

Computation of the first few terms suggests the closed form is:

$$a_n=\frac{2}{n^2-n+2}$$

As a check we may write:

$$\frac{a_{n-1}}{1+(n-1)a_{n-1}}=\frac{\dfrac{2}{(n-1)^2-(n-1)+2}}{1+(n-1)\dfrac{2}{(n-1)^2-(n-1)+2}}=\frac{2}{(n-1)^2-(n-1)+2+2(n-1)}=\frac{2}{n^2-2n+1-n+1+2+2n-2}=\frac{2}{n^2-n+2}=a_n$$

This closed form satisfies the given initial value, and the given non-linear recurrence, and so we may state:

$$a_n=\frac{2}{n^2-n+2}$$

 

[Albert1]

nice try !