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
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/10/03 21:30:38
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
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.
≥ 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.
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
利用极限存在准则证明limx趋于无穷(1/(n^6+n)^1/2+2^2/(n^6+n)^1/2+.+n^2/(n^6+
利用极限存在准则证明:limn趋向于无穷,n【1/(n^2+π)+1/(n^2+2π)+...+1/(n^2+nπ)】=
求极限 lim(n->无穷)[(3n^2-2)/(3n^2+4)]^[n(n+1)]
lim(n趋于无穷)[n(n+1)/2]/n方+3n的极限是多少?
利用极限存在准则证明当n→∞是n{[(1/(n^2+π)]+[(1/(n^2+2π)]+[(1/(n^2+3π)]+..
高数求极限 2^n*n!(/n^n) n趋于无穷?
数学极限证明:lim (n-正无穷)【(-1)^n/n^2]=0
为什么在求极限lim(1+2^n+3^n)^1/n.n-->无穷.的证明中 用夹逼定理时 (1+2^n+3^n)^1/n
求极限,lim(1+n)(1+n^2)(1+n^4)-----(1+n^2n)=?(n趋于无穷)
求极限lim((n+1)/(n2+1)+(n+2)/(n2+2)+...+(n+n)/(n2+n)),n趋近无穷
求极限lim(1/n)*[(sin(pi/n)+sin(2pi/n)+.+sin(n*pi/n)] n->无穷