Min seq

Jan 2013
96
0
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.
 
Jan 2013
96
0
I solved it! The answer is n=4! I thought that n is larger! :mrgreen: