Find general formula for $a_{n+1} = \frac{a_n}{1+n a_n}$

Question

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

$a_0=1$

Series: $1, 1/2, 1/4, 1/7, 1/11, 1/16...$ ( we ca rewrite as $a_{n+1} = \frac{1}{\frac{1}{a_n}+n}$)

By wolfram alpha answer is $\frac{2}{(n-1)^2+n+1}$

I have no idea how to get it manually.

Any ideas? Thanks

Answer 1

Hint. From $$ a_{n+1} = \frac{1}{\frac{1}{a_n}+n} $$ you get $$ \frac1{a_n}-\frac1{a_{n+1}}=-n $$ then use a telescoping sum and a standard sum.

Answer 2

Let $b_n=1/a_n$ and find the recurrence satisfied by it.