Solving non-linear recurrence relation $p_{n+1} = \frac{2p_n}{p_n + 1}, p_1 \neq 0$ Given the following recurrence relation $p_{n+1} = \frac{2p_n}{p_n + 1}, p_1 \neq 0$, I want to show that as $n \rightarrow \infty, p_n \rightarrow 1$.

By assuming that such a limit exists, and setting $p_{n+1} \approx p_n$ for large $n$, I found that $p_n$ tends to 1.

Is there a more rigorous way of doing this?

Note: I just want to show that $p_n$ tends to 1, I don't need an explicit formula for $p_n$.
Hint

Let $q_n=\frac 1 {p_n}$ to get something much more pleasant to work
If the sequence converges, it converges to $p$ such that

$$p=\frac{2p}{p+1}$$ or $$p^2-p=0.$$

But as $p_n\ne0\implies p_{n+1}\ne0$, $p=0$ is excluded, and $p=1$.

For the study of convergence, the hint by @claudeleibovici is highly recommended.