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1/n(n+2)=(1/2)[1/n-1/(n+2)]
1
1/1*3+1/2*4+1/3*5+……+1/n(n+2)
=(1/2)[1-(1/3)+(1/2)-(1/4)+(1/3)-(1/5)+……+1/n-1/(n+2)]
=(1/2)[(3/2)-3/(n+1)(n+2)]
2
(1/2)[(3/2)-3/(n+1)(n+2)]
=(3/4)-3/(n+1)(n+2)<3/4
1
1/1*3+1/2*4+1/3*5+……+1/n(n+2)
=(1/2)[1-(1/3)+(1/2)-(1/4)+(1/3)-(1/5)+……+1/n-1/(n+2)]
=(1/2)[(3/2)-3/(n+1)(n+2)]
2
(1/2)[(3/2)-3/(n+1)(n+2)]
=(3/4)-3/(n+1)(n+2)<3/4
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