1.
a(n+1)-an
=1/[(n+1)+1]+1/[(n+1)+2]+...+1/[(n+1)+(n-1)]+1/[(n+1)+n]+1/[(n+1)+(n+1)]
-[1/(n+1)+1/(n+2)+...+1/(2n)]
=1/(n+2)+1/(n+3)+...+1/(2n)+1/(2n+1)+1/(2n+2)-[1/(n+1)+1/(n+2)+...+1/(2n)]
=1/(2n+1)+1/(2n+2) -1/(n+1)
=1/(2n+1)-1/(2n+2)>0
a(n+1)>an,數(shù)列是遞增數(shù)列.
2.
證：
n=1時(shí),a1=1/(1+1)=1/2,不等式成立.
由第1問得數(shù)列是遞增數(shù)列,即n≥2時(shí),an>a1 an>1/2,不等式成立.
綜上,得an≥1/2對(duì)一切正整數(shù)n恒成立.請(qǐng)問兩式子相減之后 1/（2n+1）是哪來的？由an=1/(n+1)+1/(n+2)+...+1/(2n)知a(n+1)=1/[(n+1)+1]+1/[(n+1)+2]+...+1/[2(n+1)]a(n+1)=1/[(n+1)+1]+1/[(n+1)+2]+...+1/[2(n+1)]=1/(n+2)+1/(n+3)+...+1/(2n)+1/(2n+1)+1/(2n+2)比an多了1/(2n+1)和1/(2n+2)兩項(xiàng)，少了1/(n+1)一項(xiàng)。