✔ 最佳答案
你明白(a) part 貓朋點做吧﹐我集中精力寫靚佢個(b) part 啦
2. Given that {sn} converges to s and s > 0.
(a) Prove that there exists N ∈ Nature Numbers such that for any n ∈ Nature Numbers , if n > N then sn > 0.
Since sn →s , this means that given any ε>0 , there exits N=N(ε) such that
|sn – s| <ε whenever n > N. In particular, take ε1= s/2 , then
|sn – s| <ε whenever n > N1(ε1=s/2) and this means
-s/2 <sn – s < s/2
0< s/2 < sn < 3s/2 whenever n > N1
(b) Use ε-N definition to prove that { 3√sn} converges to 3√s.
[Hint: Recall that a^3 − b^3 = (a − b)(a^2 + ab + b^2).]
Let v= ³√s and vn= ³√sn
s=v^3 and sn=(vn)^3
From (a), we have 0< s/8 < s/2 <sn whenever n > N1(ε1=s/2) (remember s is positive) , hence
0< s/8 <sn
0< v^3/8 < (vn)^3 whenever n > N1(ε1=s/2)
0<(v/2)^3<(vn)^3
0<v/2<vn (because v and vn are positive)
Then
(v^2 + v * vn + (vn)^2 ) > (v^2 + v * v/2 + (v/2)^2 ) = 7v^2 /4...(*)
Besides,
|sn – s|
=|(vn)^3-v^3|
= |(vn – v)( v^2 + v * vn + (vn)^2) | [using hint]
= |vn – v| | v^2 + v * vn + (vn)^2 |
Rearranging
|vn – v|
= |sn – s| /| | v^2 + v * vn + (vn)^2 |
< |sn – s| / (7v^2 /4) [using (*)]
Also, we have when n>N2(ε2)
|sn – s|<ε2
[where (ε2=7v^2ε/4), notive that when ε→0, ε2→0. Because v=³√s is a finite number]
Now, given let N(ε)=max(N1,N2) , then when n>N(ε)
|vn – v|
< |sn – s| /( 7v^2 /4)
< ε2 / (ε2 /ε)
=ε
So, ³√sn→³√s
