已知数列an=n(n+1),bn=(n+1)^2,求证1/(a1+b1)+1/(a2+b2)+1/(

已知数列an=n(n+1),bn=(n+1)^2,求证1/(a1+b1)+1/(a2+b2)+1/(

(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/(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)======以下答案可供参考======供参考答案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)于是 1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn) =1/6+(1/2)(1/2-1/(n+1))=1/6+(1/4)(n-1)/(n+1)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

对的，就是这个意思

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