a1=a2=1,an+1=an+an-1,n=2,3,...xn= an+1/an.证明数列{xn}收敛于((根号5)+1)/2

an是斐波那契数列a[n+1]=an+a[n-1]a[n+1]/a[n]=1+a[n-1]/a[n]若的极限x[n]存在,收敛则lim[n->∞](a[n+1]/a[n])=lim[n->∞](a[n]/a[n-1])=xn所以xn=1+1/xn即xn^2-xn-1=0xn=(1+√5)/2 (负数略)...