作业帮若f(n)=[1/(n+1)]+[1/(n+2)]+[1/(n+3)]+``` ```+(1/2n),则f(n+1)-f(n)=
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若f(n)=[1/(n+1)]+[1/(n+2)]+[1/(n+3)]+``` ```+(1/2n),则f(n+1)-f(n)=
若f(n)=[1/(n+1)]+[1/(n+2)]+[1/(n+3)]+``` ```+(1/2n),则f(n+1)-f(n)=
若f(n)=[1/(n+1)]+[1/(n+2)]+[1/(n+3)]+``` ```+(1/2n),则f(n+1)-f(n)=
[1/（2n+1）]-[1/(2n+2)]
f(n+1)=1/(n+2) + 1/(n+3) +……+ 1/(2n+1) + 1/(2n+2)
所以两者之差=1/(2n+1) + 1/(2n+2) - 1/(n+1)
1/(2n+1)-1/(2n+2)
f(n)=1/(n+1)+1/(n+2)+1/(n+3)+ ```+1/(2n)
f(n+1)= 1/(n+2)+1/(n+3) +… +1/(2n) +1/(2n+1) + 1/(2n+2)
则f(n+1)-f(n)=1/(2n+1)+1/(2n+2)-1/(n+1)
=1/(2n+1)+1/(2n+2)-2/(2n+2)
=1/(2n+1)-1/(2n+2)
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