a(n+1)²-an²=2
∴數(shù)列{an²}為等差數(shù)列,公差d=2
a1²=1
∴an²=1+(n-1)×2=2n-1
又∵an＞0
∴an=√(2n-1)求證:1/a1+1/a2+。。。+1/an<=根號(hào)下2n-11/a1+1/a2+……+1/an=1+1/√3+1/√5+……+ 1/√(2n-1)=1+2/(2√3)+2/(2√5)+……+ 2/[2√(2n-1)]≤1+2/(√1+√3)+2/(√3+√5)+……+ 2/[√(2n-3)+√(2n-1)]=1 +(√3-√1)+ (√5-√3) + ……+ [√(2n-1.)-√(2n-3)]=√(2n-1)