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If-n-Z-prove-that-1-2-1-1-3-2-1-4-3-1-n-1-n-lt-2-




Question Number 113070 by ZiYangLee last updated on 11/Sep/20
If n∈Z^+ , prove that  (1/(2(√1)))+(1/(3(√2)))+(1/(4(√3)))+...(1/((n+1)(√n)))<2
IfnZ+,provethat121+132+143+1(n+1)n<2
Answered by 1549442205PVT last updated on 11/Sep/20
We have (1/((n+1)(√n)))=(√n).(1/((n+1)n))  =(√n)((1/n)−(1/(n+1)))=(√n)((1/( (√n)))+(1/( (√(n+1)))))((1/( (√n)))−(1/( (√(n+1)))))  =(1+(1/( (√(n+1)))))((1/( (√n)))−(1/( (√(n+1)))))  Since (1+(1/( (√(n+1)))))<2,so [(1/((n+1)(√n)))]  <2((1/( (√n)))−(1/( (√(n+1))))).Hence,give n=1,2,...  we get  (1/(2(√1)))=(1/2)<2((1/1)−(1/( (√2)))),(1/(3(√2)))<2((1/( (√2)))−(1/( (√3)))),  ...,(1/((n+1)(√n)))<2((1/( (√n)))−(1/( (√(n+1)))))  Adding up n above inequalities we get  LHS=(1/(2(√1)))+(1/(3(√2)))+(1/(4(√3)))+...(1/((n+1)(√n)))<  2[(1/( (√1)))−(1/( (√2)))+(1/( (√2)))−(1/( (√3)))+...+(1/( (√n)))−(1/( (√(n+1)))))  =2(1−(1/( (√(n+1)))))<2  Therefore,we have   (1/(2(√1)))+(1/(3(√2)))+(1/(4(√3)))+...(1/((n+1)(√n)))<2(q.e.d)
Wehave1(n+1)n=n.1(n+1)n=n(1n1n+1)=n(1n+1n+1)(1n1n+1)=(1+1n+1)(1n1n+1)Since(1+1n+1)<2,so[1(n+1)n]<2(1n1n+1).Hence,given=1,2,weget121=12<2(1112),132<2(1213),,1(n+1)n<2(1n1n+1)AddingupnaboveinequalitieswegetLHS=121+132+143+1(n+1)n<2[1112+1213++1n1n+1)=2(11n+1)<2Therefore,wehave121+132+143+1(n+1)n<2(q.e.d)

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