分析:由于\(2a_na_{n+1}=a_n^2+1\),则可知\(a_{n+1}=\cfrac{a_n^2+1}{2a_n}=\cfrac{1}{2}(a_n+\cfrac{1}{a_n})\),
则\(b_{n+1}=\cfrac{a_{n+1}-1}{a_{n+1}+1}=\cfrac{\cfrac{1}{2}(a_n+\cfrac{1}{a_n})-1}{\cfrac{1}{2}(a_n+\cfrac{1}{a_n})+1}=\cfrac{(a_n-1)^2}{(a_n+1)^2}=b_n^2\),
又由于\(a_1=2\),\(b_1=\cfrac{a_1-1}{a_1+1}=\cfrac{1}{3}\),故\(b_2=b_1^2=(\cfrac{1}{3})^2\),\(b_3=b_2^2=(\cfrac{1}{3})^4\),\(b_4=b_3^2=(\cfrac{1}{3})^8\),故数列\(\{b_n\}\)为递减数列。故选\(D\)。
原文:https://www.cnblogs.com/wanghai0666/p/10696741.html