(1)令n=1得,a1+a1=2
a1=1
a（2n-1）=2n-1
a（n）=n
(2)f(n)=s(2n)-s(n)=1/(n+1)+1/(n+2)+^^^+1/(2n)
f(n+1)=s(2n+2)-s(n+1)=1/(n+2)+1/(n+3)+^^^+1/(2n)+1/(2n+1)+1/(2n+2)
f(n+1)-f(n)=1/(2n+1)+1/(2n+2)-1/（n+1)>1/(2n+2)+1/(2n+2)-1/（n+1)=0
f(n+1)>f(n)
(3)当n=1时,f(n)的最小值为2
g(x)