无穷)n^2[1/(n^2+1)^2+2/(n^2+2)^2+...+n/(n^2+n)^2]=1/

发布时间:2021-02-20 16:47:09

无穷)n^2[1/(n^2+1)^2+2/(n^2+2)^2+...+n/(n^2+n)^2]=1/2

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1.n^2[1/(n^2+1)^2+2/(n^2+2)^2+...+n/(n^2+n)^2]
≥ n^2[1/(n^2+n)^2+2/(n^2+n)^2+...+n/(n^2+n)^2]
= n^2[1+2+...+n]/[(n^2+n)^2]
= n^2[n(n+1)/2]/[(n^2+n)^2]
= (1/2)[n^4+n^3]/[n^4+2n^3+n^2]
(1/2)[n^4+n^3]/[n^4+2n^3+n^2]中令n->∞,极限是1/2
2.n^2[1/(n^2+1)^2+2/(n^2+2)^2+...+n/(n^2+n)^2]
≤ n^2[1/(n^2+1)^2+2/(n^2+1)^2+...+n/(n^2+1)^2]
= n^2[1+2+...+n]/[(n^2+1)^2]
= n^2[n(n+1)/2]/[(n^2+1)^2]
= (1/2)[n^4+n^3]/[n^4+2n^3+1]
(1/2)[n^4+n^3]/[n^4+2n^3+1]中令n->∞,极限是1/2
根据夹逼定理(准则),知道极限存在,并且极限是1/2.
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