极限证明*(证完再加分)

来源:百度知道 编辑:UC知道 时间:2024/05/24 00:42:32
(1)
lim (1^k+2^k+...+n^k)/n^k+1=1/k+1
n-无穷

(2)
证:
[1/m(a1^1/n+a2^1/n+a3^1/n+...+an^1/n)]^n
=(a1*a2*...*an)^1/m

*注:第二题已有一定证明步骤

令bn=原式=1-1/m{[a1^1/n-1]+[a2^1/n-1]+...+[an^1/n]}^n
=(1+cn)^n

证出任何一问小弟会感激不尽*****

lim (1^k+2^k+...+n^k)/n^(k+1)
=lim(1/n)*{(1/n)^k+(2/n)^k+...+(i/n)^k+..+(n/n)^k}因为n趋于无穷
上式就相当于积分x^kdx,在0到1的积分
x^k的原函数为x^(k+1)/k+1,所以等于1^(k+1)/k+1-0=1/k+1
问题得证

(1)的结论不成立。当k=1时:
lim (1^k+2^k+...+n^k)/n^k+1
n-无穷
=lim (1+2+...+n)/n+1
n-无穷
=lim (1+2+...+n)/n+1
n-无穷
=lim (n/2)≠1/2.
n-无穷

1.
1)令Ak = (1^k+2^k+...+n^k)/n^(k+1)
用数学归纳法证的 Ak > 1/k+1 和 (k+2)*A(k+1) <= (k+1)*Ak
2)夹逼定理
1/(k+1) < Ak < A(k-1) < A1
以下n-无穷
当k = 1时
lim A1 = 1/(1+1) = 1 /(k+1);
假设lim Ak = 1 /(k+1)成立;
当k = k + 1时
有lim A(k+1) <lim (k+1)*Ak /k+2) = (k+1)/(k+2)limAk = 1/(k+2);
所以当k = k + 1时成立
所以lim A(k+1) < limAk < lim A1 = 1 / (k+1);
lim Ak > lim1/(k+1) =
1/(k+1)
lim Ak = 1/(k+1)

看不懂