Find the limit of $(x_n)$ defined by $x_{n+1}=c_nu(x_n)$ Find $\displaystyle \lim_{n\rightarrow \infty }x_n$ :
$$\left\{\begin{matrix}x_1=a>0\\ \\ x_{n+1}=\frac{2x_n\cdot \cos\left(\frac{\pi}{2^n+1}\right)}{x_n+1}\end{matrix}\right.$$
I have tried that :
Let $a_n=\dfrac{1}{x_n}$ . So :
$$a_{n+1}=\frac{1}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}.a_n+\frac{1}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}$$
So I tried to have geometric series: 
By let :
$$a_{n+1}+f(n+1)=\frac{1}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}(a_n+f(n))$$
So we must find one $f(n)$:
$$\frac{f(n)}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}-f(n+1)=\frac{1}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}$$
As $$f(n)-f(n+1)\cdot 2\cos\frac{\pi}{2^{n+1}}=1$$
Can you give me the way to find one $f(n)$ sastisfied that ; or anyone has nice way to solve this problem

An incomplete answer that assumes the existence of the limit:
Since $\cos\frac{\pi}{2^n+1}$ goes to one for $n\to\infty$, the limit $x_\infty=\lim_{n\to\infty}x_n$ will, if it exists, satisfy the fixed-point equation $x_\infty=2x_\infty/(x_\infty+1)$. The only solutions of this equation are $0$ and $1$.
The derivative of $x\mapsto 2x\cos\frac{\pi}{2^n+1}/(x+1)$ at $x=0$ equals $2\cos\frac{\pi}{2^n+1}$ which is greater than one for $n>1$. Fixed-point theory then tells us that $x=0$ can not be the limit of your recursion. The derivative at $x=1$, in contrast, equals $1/2\cos\frac{\pi}{2^n+1}$ which is always positive and less than unity.
