求证: 16<1/√1+1/√2+...+1/√80<17.

求证: 16<1/√1+1/√2+...+1/√80<17.

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2016-08-26



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1/√1+1/√2+1/√3……1/√80 =1/√1+2/2√2+2/2√3……2/2√80 =1/√1+2/(√2+√2)+2/(√3+√3)……2/(√80+√80) <1/√1+2[1/(√2+1)+1/(√3+√2)……1/(√80+√79)] =1+2[(√2-1)+(√3-√2)+……+(√80-√79)] =1+2(-1+√80) <1+2(-1+√81) =17 所以1/√1+1/√2+1/√3……1/√80<17 1/√1+1/√2+1/√3……1/√80 =1/√1+1/√2+2/2√3……2/2√80 =1/√1+1/√2+2[1/(√3+√3)+……+1/(√80+√80)] >1/√1+1/√2+2[1/(√3+√4)+……+1/(√80+√81)] =1/√1+1/√2+2[(√4-√3)+……+(√81-√80)] =1+1/√2+2(-√3+√81) 比较1+1/√2+2(-√3+√81)和16的大小 1+1/√2+2(-√3+√81)-16 =3+1/√2-2√3 (3+1/√2)²=19/2+6/√2 (2√3)²=12 (12-19/2)²=25/4 (6/√2)²=18>25/4 所以12-19/2<6/√2 19/2+6/√2>12 即3+1/√2-2√3>0 所以1/√1+1/√2+1/√3……1/√80>1+1/√2+2(-√3+√81)>16 所以16<1/√1+1/√2+1/√3……1/√80<17

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