求证(1/n+1)+(1/n+2)+……(1/3n)>(5/6)[n>=2]

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求证(1/n+1)+(1/n+2)+……(1/3n)>(5/6)[n>=2]

求证(1/n+1)+(1/n+2)+……(1/3n)>(5/6)[n>=2]
求证(1/n+1)+(1/n+2)+……(1/3n)>(5/6)[n>=2]

求证(1/n+1)+(1/n+2)+……(1/3n)>(5/6)[n>=2]
构造函数:
f(n)=(1/n+1)+(1/n+2)+.+(1/3n),n≥2
∴f(n+1)=(1/n+2)+.+(1/3n)+(1/3n+1)+(1/3n+2)+(1/3n+3)
∴f(n+1)-f(n)=(1/3n+1)+(1/3n+2)+(1/3n+3)-(1/n+1)
>(1/3n+3)+(1/3n+3)+(1/3n+3)+(1/n+1)=0
∴f(n)在N+上是单调递增的
故f(n)>f(2)=1/3+1/4+1/5+1/6=57/60>5/6