100分求高数难题...写的好追分..百思不得其解的题目...1.An=[∑1/(2*(1+2+3...+k))]^n∑是n=1到n则n->∞limAn=2.f(x)=√(x+2)-√(x+1)+√x求ck使x->∞时f(x)~c/(x^k)答案c=-1/4k=3/23.若f(x)在(-∞,∞)上二阶

Question

100分求高数难题...写的好追分..
百思不得其解的题目...
1.
An=[∑1/(2*(1+2+3...+k))]^n ∑是n=1到n
则n->∞ lim An=______
2.
f(x)=√(x+2)-√(x+1)+√x
求c k 使x->∞ 时 f(x)~c/(x^k)
答案c=-1/4 k=3/2
3.若f(x) 在(-∞,∞)上二阶可导 Mi=Max|f(x)的i阶导数|

Answer 1

1. 1+2+...+k=k(k+1)/22*(1+2+...+k)=k(k+1)∑1/(2*(1+2+3...+k))=(1/1*2)+(1/2*3)+(1/3*4)+...+(1/n(n+1))=[(1/1)-(1/2)]+[(1/2)-(1/3)]+[(1/3)-(1/4)]+...+[(1/n)-(1/(n+1))]=1-(1/(n+1))=n/(n+1)An=[n/(n+1)]^nl...