Solving recursion $f(n+1)=\frac n{n+2}f(n)$

Question

How can I find an expression for $$f(n), n \in \mathbb{N}$$ if $$f(n+1)=\frac{n}{n+2}\cdot f(n)$$ and $$f(1) = 999$$

I'm used to solve simple homogeneous recursion relations by the chacarcteristic equation, but it seems impossible in this situation. What can I do?

Thank you for your attention.

Answer 1

Multiplying the both sides of $$f(n+1)=\frac{n}{n+2}f(n)$$ by $(n+1)(n+2)$ gives $$(n+1)(n+2)f(n+1)=n(n+1)f(n)$$ Here, if we define $g(n)=n(n+1)f(n)$, then we get $$g(n+1)=g(n)$$

Since $$g(n)=g(1)=1\times 2\times f(1)=2\times 999=1998$$

we get $$f(n)=\frac{g(n)}{n(n+1)}=\frac{1998}{n(n+1)}$$