斐波那契1 1 2 3 5 8.怎么证明前后比值无限接近0.618.

斐波那契数列通项公式F(n)=(1/√5)*{[(1+√5)/2]^n - [(1-√5)/2]^n}F(n)/F(n+1)= {[(1+√5)/2]^n - [(1-√5)/2]^n}/{[(1+√5)/2]^(n+1) - [(1-√5)/2]^(n+1)}lim(n→∞）F(n)/F(n+1)=lim(n→∞）{[(1+√5)/2]^n - [(1-√5)/2]^n}/{[(1+√5)/2]^(n+1) - [(1-√5)/2]^(n+1)}=lim(n→∞）[(1+√5)/2]^n /[(1+√5)/2]^(n+1)=1/[(1+√5)/2]=(√5-1)/2=0。618