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