Question

利用极限存在准则证明lim(n—>无穷)n^2[1/(n^2+1)^2+2/(n^2+2)^2+...+n/(n^2+n)^2]=1/2

如题，要过程详解

Answer 1

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.

Answer 2

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Answer 3

妹的你学数分的吧....

追问

求解释。 要过程

追答

不做数分题好多年.....太蛋疼了....不好意思灌水了...