Solve recurrence $a(n+1)= \frac{a(n)}{a(n)+2}$ with $a(0)=1$

Question

Solving recurrence

$a(n+1)= \cfrac{a(n)}{a(n)+2}$

with $a(0)=1$

Do I have to make a replace? Can someone help with initial steps?

Thanks.

Answer 1

Let $b(n)=\frac{1}{a(n)}$. Then $$b(n+1)=1+2b(n);\,b(0)=1$$

Solve this for $b(n)$.