Olympic question about functions

Question

I have been trying to solve the following question:

Let $f: \left \{ 1,2,... \right \}\rightarrow \mathbb{R}$ be a function such that $f(n) - f(n+1) = f(n)f(n+1)$. If $f(2020) = \frac{1}{4040}$, find $f(1)$

The answer is f(1) = $\frac{1}{2021}$.

Using a simple algebraic manipulation, I found that if $f(2020) = \frac{1}{4040}$ then $f(2019) = \frac{1}{4039}$, what is interesting for the problem considering the answer...

How can i progress? Or there a trick way in these question

Answer 1

Equivalently we can show that $f(n+1) = \frac{f(n)}{f(n) + 1}$

By recursive application, easy to show $f(n+k) = \frac{f(n)}{k * f(n) + 1}$

Now let $n=1$ and $k=2019$, then

$$f(2020) = \frac{f(1)}{2019 * f(1) + 1} = \frac{1}{4040}$$ $$\Rightarrow 4040*f(1) = 2019 * f(1) + 1 \Rightarrow f(1) = \frac{1}{2021}$$