We need to show that {an}is bounded above by 3 and monotonic increasing
a1=-2<3
Assume ak=√(6+ak-1)<3
then ak+1=√(6+ak)<3 ⇒for all n=k+1
By M.I {an}<3
a1=-2 a2=2 a1<a2 when n=1
Assume true for n=k ak<ak+1⇒ak+1/ak>1
then ak+2/ak+1=√(6+ak+1)/√(6+ak)>1 for n=k+1
By M.I an+1>an
{an}is monotonic increasing
let liman=liman+1=L as n→∞
L=√(6+L) L=3作者: 23909504 時間: 2019-3-14 10:27
圖片附件: 微積分.png (2019-3-13 21:13, 47.71 KB) / 下載次數 18