Showing that $\lim_{n \to \infty} x_n = 0$ Let $(x_n)^{\infty}_{n=1}$ a sequence, such :
$$x_1=1 \; \text{and} \; x_{n+1}=\frac{n}{n+1} (x_n)^2, \; n \in \mathbb{N}$$
Show that $x_n \xrightarrow{ n \to \infty} 0$.
I have no idea how I would prove this? I mean the sequence is clearly decreasing but I don't know where to go from there?

For each $n\in\mathbb N\setminus\{1\}$, $x_n<{x_{n-1}}^2$. Since $x_2=\frac12$, it follows that $x_n\leqslant\frac1{2^{n-1}}$ for each $n\in\mathbb N$. Can you take it from here?
