已知数列an=n(n+1),bn=(n+1)^2,求证1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)
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已知数列an=n(n+1),bn=(n+1)^2,求证1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)
已知数列an=n(n+1),bn=(n+1)^2,求证1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)
已知数列an=n(n+1),bn=(n+1)^2,求证1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)
(an+bn)=n(n+1)+(n+1)^2
=(n+1)(2n+1)>2n(n+1)
1/(an+bn)=1/(n+1)(2n+1)
=2/(2n+1)-2/(2n+2)
从第二项开始放缩,即1/(an+bn)=1/(n+1)(2n+1)< 1/2n(n+1)=1/2（1/n-1/n+1）
1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)<
2/3-2/4+1/2(1/2-1/3+1/3-1/4+.+1/n-1/n+1
=1/6+1/2(1/2-1/n+1)
由(an+bn)=n(n+1)+(n+1)^2 =(n+1)(2n+1)>2n(n+1)
故得1/(an+bn)=1/(n+1)(2n+1)<1/(2n(n+1))=(1/2)(1/n-1/(n+1))
于是
1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)
<1/6+(1/2)(1/2-1/3+1/3-1/4+...+1...
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由(an+bn)=n(n+1)+(n+1)^2 =(n+1)(2n+1)>2n(n+1)
故得1/(an+bn)=1/(n+1)(2n+1)<1/(2n(n+1))=(1/2)(1/n-1/(n+1))
于是
1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)
<1/6+(1/2)(1/2-1/3+1/3-1/4+...+1/n-1/(n+1))
=1/6+(1/2)(1/2-1/(n+1))=1/6+(1/4)(n-1)/(n+1)<1/6+1/4=5/12.
利用非初等方法还可以改进这个结果,
由an+bn=n(n+1)+(n+1)^2=(n+1)(2n+1)，
1/(an+bn)=1/((2n+2)(2n+1))=2(1/(2n+1)-1/(2n+2))，
故得1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)
=2(1/3-1/4+1/5-1/6+....+1/(2n+1)-1/(2n+2))
1-1/2+1/3-1/4+....=ln2,得1/3-1/4+....=ln2+1/2-1=ln2-1/2.
1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)+...
=2(ln2-1/2)=0.386294361...
该级数是递增的,故对所有n原式<2(ln2-1/2)=0.386294361...,较5/12=0.4166666...上界要小,不过1楼做得很好.
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