# Thread: Advanced Calculus - Convergence Sequence Problem

1. ## Advanced Calculus - Convergence Sequence Problem

I had been trying to solve these problems for the entire weekend, and I just can't think of anything that would work, please help!!!

#1. Let c be a real number. Prove that there is an n of Natural Number such that 2^n > c

My solution: 2^n is not bounded so I can't use that to prove c can be a lower bound, then I tried to use the archmedian principle and it still won't help.

#2. Let S = (k/2^n : k within Natural Number, n without Natural Number). Let 0 <= a < b. Prove that there is an x within S such that a < x < b.

My solution: I think this is to prove that S is dense, but the 2^n is really messing up my pace here, I can't prove this problem as like other "dense" set...

#3. Let a1 = 4. Define an+1 = an/2 + 2/an , for n = 1,2,3,...

a) Evaluate
a2, a3, a4.
b) Prove that if an > 2, then an+1 > 2. use induction to prove that an > 2 for all n within the set of Natural Numbers.
c) Prove that
an+1 <= an for all n within the set of Natural Numbers.
d) Is {
an} bounded? If so, find A,B of R such that A <= an <= B for all n within N.
e) Does lim (n->inf)
an exist? Why?

My solution: I'm currently still reading the text in attempt to find the answer.

Finally, I understand that these problems are hard and would reqire some work to finish, but if you can as much as give me some hints I would really appreicate it!!!

Thank you!

2. Originally Posted by tttcomrader

#1. Let c be a real number. Prove that there is an n of Natural Number such that 2^n > c
Suppose otherwise. Then 2^n, n=1, 2, .. is an increasing sequence which
is bounded above, and so by the least upper bound axiom has a least upper
bound 0<C<=c.

So we have for all n in N+:

2^n <= C,

so we also have:

2^(n+1) = 2.2^n <= C,

so:

2^n <= C/2.

but C/2 < C, contradicting the assumption that C was a least upper bound.

RonL

3. Originally Posted by tttcomrader

#3. Let a1 = 4. Define an+1 = an/2 + 2/an , for n = 1,2,3,...

a) Evaluate
a2, a3, a4.
b) Prove that if an > 2, then an+1 > 2. use induction to prove that an > 2 for all n within the set of Natural Numbers.
c) Prove that
an+1 <= an for all n within the set of Natural Numbers.
d) Is {
an} bounded? If so, find A,B of R such that A <= an <= B for all n within N.
e) Does lim (n->inf)
an exist? Why?

My solution: I'm currently still reading the text in attempt to find the answer.

Finally, I understand that these problems are hard and would reqire some work to finish, but if you can as much as give me some hints I would really appreicate it!!!

Thank you!
Use the Bolzano-Weierstass Theorem.

Every Monotone Increasing sequence with an upper bound has a limit.

4. Originally Posted by tttcomrader

#2. Let S = (k/2^n : k within Natural Number, n without Natural Number). Let 0 <= a < b. Prove that there is an x within S such that a < x < b.
I am not sure I understand the question.
The largest element in S is when n=0 that is, k.
But if a>k then there is no way that you can find an element in S such as,
a<x<b.

Also, the word "dense" was used false here. Between any two elements a,b in an ordered set we have infinitely many,
a<x<b.

5. I had finished most of the problem, but I'm not sure if they are all correct, please check!

1. Let c be within the set of all Real Numbers. Prove that there is an n within the set of Natural Numbers such that 2^n > c.

My solution:
Let S = { n E N* : 2^n > n }
*n within the set of Natural Numbers, I don't know how to type it in here.

Proof S is inductive:
a) 1 E N, 2^1 = 2 > 1 , 1 E S.
b) Let k E N, then k E N, 2^k > k, then (k+1) E N, 2^(k+1) = 2^k (2) > 2k > k+1
c) Therefore, S is inductive.
d) By the archimedean property, there is an N within the set of natural number such that N > c for all c within the set of Real Numbers. Let n >= N, so 2^n > n > c. Q.E.D.

2. Let S = { (k/2^n)* : k E N, n E N}. Let 0 <= a < b. Prove that there is an x E S such that a < x < b.
*In response to the question posted by ThePerfectHacker, the max k is not 0 as k can be any natural number and not depended on n, therefore k can be infinity.

My solution:
a) Let S = { k E N : (k/2^n) < b }, and let c* = b - a.
*from the first question 2^n > c
b) Let M be the max of S, then (M/2^n) <= b
c) Then (M+1)/(2^n) is not in S, and is >= b
d) Then M/2^n >= b - 1/2^n > a
e) So: b > M/2^n > a Q.E.D.