![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
求极限limx→∞[1^2/(n^3+1)+2^2/(n^3+2)+……+n^2/(n^3+n)]
如题:limx→∞[1^2/(n^3+1)+2^2/(n^3+2)+……+n^2/(n^3+n)]详解...
如题:limx→∞[1^2/(n^3+1)+2^2/(n^3+2)+……+n^2/(n^3+n)]详解
展开
1个回答
展开全部
(1^2+2^2+...+n^2)/(n^3+n)<=1^2/(n^3+1)+2^2/(n^3+2)+……+n^2/(n^3+n)
<=(1^2+2^2+...+n^2)/(n^3+1),
——》n(n+1)(2n+1)/6(n^3+n)<=1^2/(n^3+1)+2^2/(n^3+2)+……+n^2/(n^3+n)
<=n(n+1)(2n+1)/6(n^3+1),
limn→∞ (n+1)(2n+1)/6(n^2+1)=limn→∞ (1+1/n)(2+1/n)/6(1+1/n^2)=2/6=1/3,
limn→∞ n(n+1)(2n+1)/6(n^3+1)=limn→∞ (1+1/n)(2+1/n)/6(1+1/n^3)=2/6=1/3,
——》limn→∞ (n+1)(2n+1)/6(n^2+1)<=原式<=limn→∞ n(n+1)(2n+1)/6(n^3+1),
由夹逼定理知:
原式=1/3。
<=(1^2+2^2+...+n^2)/(n^3+1),
——》n(n+1)(2n+1)/6(n^3+n)<=1^2/(n^3+1)+2^2/(n^3+2)+……+n^2/(n^3+n)
<=n(n+1)(2n+1)/6(n^3+1),
limn→∞ (n+1)(2n+1)/6(n^2+1)=limn→∞ (1+1/n)(2+1/n)/6(1+1/n^2)=2/6=1/3,
limn→∞ n(n+1)(2n+1)/6(n^3+1)=limn→∞ (1+1/n)(2+1/n)/6(1+1/n^3)=2/6=1/3,
——》limn→∞ (n+1)(2n+1)/6(n^2+1)<=原式<=limn→∞ n(n+1)(2n+1)/6(n^3+1),
由夹逼定理知:
原式=1/3。
更多追问追答
追问
开头是怎么个思路,
追答
各分式的分母不同不能直接加,就设想换同分母来便于计算,
再考虑夹逼定理。
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询