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,数列是递增数列.
2.
证：
n=1时,a1=1/(1+1)=1/2,不等式成立.
由第1问得数列是递增数列,即n≥2时,an>a1 an>1/2,不等式成立.
综上,得an≥1/2对一切正整数n恒成立.