Why this recursively defined sequence of real numbers converges to -Pi? Remy J. Cano in his private email described the sequence of real numbers, recursively defined as
$$a(n) = a(n-1)+\frac{2 \cdot \cos(\frac{a(n-1)}{2})}{2 \cdot \sin(\frac{a(n-1)}{2})-1},a(0)=0$$
This sequence converges to $-\pi$
that is for $n \rightarrow  \infty $
$ a(n) \rightarrow  -\pi$
Why this recursively defined sequence of real numbers converges to $-\pi$ ?
 A: Look at: $$T(x) = x+\frac{2 \cdot \cos\left(\frac{x}{2}\right)}{2 \cdot \sin\left(\frac{x}{2}\right)-1}$$
We know that $a_1=-2$. When $x \in X\equiv(-2\pi,0)$, $T(x)\in X$ as well. Moreover, since $T(x)$ has a continuous derivative in $X$, and: 
$$|T'(x)|<1\ \mid\  x\in X,$$
$T(x)$ is a contraction mapping.
Thus, by the Banach Fixed Point Theorem, there exists a single stationary point. Since $x=-\pi$ is such a point, the series converges to $x=-\pi$.
A: The main reason is that $\cot(x)$ function have zeros at odd multiples of $\pi$.
It is relatively similar but not the same than the original function I mailed. Almost any other thing (real number) used instead of your choosen $a_{0}$ will behave in the same way, since there are infinitely many multiples of $\pi$, therefore you always can set by convention any stop point (consisting in: To reach some desired precision), getting an approximation to an odd integer multiple of $\pi$. Computer Software with smart capabilities for dealing with trigonometric functions like PARI, have already pre-computed $\pi$ to any allowed precision, so it is not difficult to realize the described convergence by inspection dividing $a_{n}$ between $\pi$. (From: R. J. Cano, Physics student and CS amateur)
