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(1)bn≤0时bn0时,a1=1,b1^2≤b1-b2<b1,得b1<1,即有b1<a1设bn<an 成立,由bn^2≤bn-bn+1 得bn+1≤bn- bn^2=-(1/2-bn)^2+1/4<-(1/2-an )^2+1/4=an-an^2=1/n-1/n^2=(n-1)/n^2<1/(n+1)即有bn+1<an+1(2)b1+b2+b3+……+b(2^n-1)<1+1/2+1/3+....+1/(2^n-1)1/2+1/3<1,1/4+/1/5+1/6+1/7<1,.....,1/2^(n-1)+1/[2^(n-1)+1]+....+1/(2^n-1)<2^(n-1)*[1/2^(n... |
