Showing that $\lim_{n \to \infty} x_n = 0$

Question

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?

Answer 1

The sequence is decreasing and bounded below (by zero), and therefore convergent: $a = \lim_{n\to \infty} x_n$ exists.

Now consider the identity $$x_{n+1}=\frac{n}{n+1} x_n^2 \,.$$ For $n \to \infty$, the left-hand side converges to $a$, and the right-hand side converges to $a^2$.

It follows that $a = a^2$, so that $a=0$ or $a=1$ are the only possible limits. The latter is not possible because $x_n \le \frac 12$ for $n \ge 2$.

Answer 2

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?