# Min seq

#### yo79

Let $$\displaystyle (x_n)_{n\ge 1}$$ be a sequence defined by $$\displaystyle x_1=1$$ and $$\displaystyle x_{n+1}=\frac{x_n^2+2}{2x_n}$$. It is known that $$\displaystyle (x_n)$$ converges increasing to $$\displaystyle \sqrt{2}$$. Which is the smallest n such that $$\displaystyle 0,4143 \ge \{ x_n \} \ge 0,4142$$, where {x}=x-[x], and [x] is the largest integer less or equal to x.

#### yo79

I solved it! The answer is n=4! I thought that n is larger! :mrgreen: