Finding 1996'th term for given recurrence relation

Question

I've been given the sequence $a_n$ defined by the recursion relation: $$a_{n+1}=\frac{a_n}{1+na_n}\qquad a_1=1$$

and have been tasked to find $a_{1996}$. How would I go about that?

I have a basic knowledge about recursions.

I don't know how to isolate $a_n$ since it appears twice in the equation.

Anybody have a clue how to proceed?

Answer 1

$$ \frac1{a_{n+1}}=\frac1{a_n}+n $$

let $b_n = \frac1{a_n}$ $$ b_{n+1}=b_n + n $$ Using recursion, $$ b_1 = 1\\ b_2 = 1+1\\ b_3 = 1+1+2\\ \vdots\\ b_n = 1+\sum_{n=0}^{n-1} n = 1+\frac{n(n-1)}{2} $$ Then, $$ a_n = \frac{2}{2+n(n-1)} $$