(1) na(n+1)=(n+1)an a(n+1)/an=(n+1)/n a2/a1=2/1 a2/a2=3/2 a4/a3=4/3 …… an/a(n-1)=n/(n-1) a(n+1)/an=(n+1)/n 将上列式子相乘得 a(n+1)/a1=2(3/2)(4/3)(5/4)……[n/(n-1)][(n+1)/n] a(n+1)=n+1 an=n Tn=1/(1*2)+1/(2*3)+1/(3*4)+……+1/[(n-1)n]+1/[n(n+1)] =(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+[1/(n-1)-1/n]+[1/n-1/(n+1)] =1-1/(n+1) =n/(n+1) (2)Tn=n/(n+1)<t^2-0.5 t^2>n/(n+1)+0.5=(n+1-1)/(n+1)+0.5=1.5-1/(n+1) 1<=n<=9 t^2>=1.5-1/(1+1)=1 t<-1或t>1 第二问不敢保证