Does this recurrence really converge? Iterating by computer $x_0=0$, $x_n=\frac{{x_{n-1}}^2}{2}-1$ I got that it seems to converge $\lim_{n\to\infty}x_n = 1-\sqrt3$.
Sadly I have tried everything to prove if that is true or false without success, could you help me?, thank you.
 A: You are iterating the function $f(x) = x^2/2 - 1$.  $p = 1 - \sqrt{3}$ is a fixed point (i.e. $f(p) = p$), and $f'(p) = 1-\sqrt{3}$.  Note that $-1 < f'(p) < 0$, and $-1 < f'(x) < 0$ for $-1 < x < 0$.  Thus if $-1 < x < 0$ the Mean Value Theorem implies $|f(x) - p| < |x - p|$.  As soon as $-1 < x_n < 2p+1= 3-\sqrt{3}$ (an interval symmetric around $p$), we can conclude that $|x_n-p|$ will be decreasing from that point on.  In this case that is true already for 
$x_2 = -1/2$.
A: First of all we can observe that $\epsilon := -7/8 < x_n \leq 0$ for all $n > 3$ if $x_0 = 0$ and so
\begin{equation}
|x_{n+1} - x_{n}| = \frac{1}{2}|x^2_{n-1} - x^2_{n-2}| = \frac{1}{2}|x_{n-1} + x_{n-2}||x_{n-1} - x_{n-2}| < \epsilon|x_{n-1} - x_{x-2}|.
\end{equation}
Therefore we can see that $\{x_n\}$ is a Cauchy sequence in $\mathbb{R}$ hence converges, (it is not hard to see that $|x_n - x_m| \rightarrow 0$ for large $n,m$. To get the limit we can find the fixed point from $x = x^2/2 - 1 \implies x = 1 - \sqrt{3}$ (pick the root such that $-1 \leq x \leq 0$).
