cho dãy (un) xác định bởi
$\left\{\begin{matrix} u_{1}=a>2 & & \\ u_{n+1}= u_{n}^{2}-2 & \end{matrix}\right.$
Tính
$\lim_{n \to \infty }(\frac{1}{u_{1}}+\frac{1}{u_{1}u_{2}}+\frac{1}{u_{1}u_{2}u_{3}}+...+\frac{1}{u_{1}u_{2}u_{3}...u_{n}})$
Đặt $ a = x +\frac{1}{x} ( ax > 0 ) $
thì
$u_1 = x +\frac{1}{x} $
$u_2 = (x +\frac{1}{x})^2 -2 = x^2 +\frac{1}{x^2} $
quy nạp chứng minh được
$u_n = x^{2^{n-1}} + \frac{1}{x^{2^{n-1}}} $
Gọi $ M = \frac{1}{u_{1}}+\frac{1}{u_{1}u_{2}}+\frac{1}{u_{1}u_{2}u_{3}}+...+\frac{1}{u_{1}u_{2}u_{3}...u_{n}}$
Thì
$M = \frac{x}{x^2 + 1} + \frac{x^3}{(x^2+1)(x^4+1)} + ..+ \frac{x^{2^n-1}}{(1+x^2)(1+x^4)..(1+x^{2^n})}$
$M + \frac{x-x^{2^{n+1}-1}}{x^{2^{n+1}}-1} = 0$
$\Rightarrow M = \frac{-x+ x^{2^{n+1}-1}}{x^{2^{n+1}}-1} = 0 $
$\lim_{n \to \infty } M $ $= \lim_{n \to \infty } \frac{\frac{1}{x} - \frac{1}{x^{2^{n+1}-1}}}{1-\frac{1}{x^{2^{n+1}}}}$$=\frac{1}{x}$