Is There an N Where n > N Implies Sn > a in a Convergent Sequence?

In summary: And that shows that there exists N such that if n>N, S[sub]n> a.In summary, the conversation discusses proving the existence of a specific number N in a convergent sequence, where the limit of the sequence is greater than a. The use of the sequence definition of limit and choosing a specific epsilon value is mentioned, and the method of using the distance between a and b to make a specific choice for epsilon is suggested. The conversation also brings up the idea of choosing epsilon to be less than the distance between a and b, and how this can be used in the proof. Finally, the conversation concludes with a summary of the method used to prove the existence of N in this scenario.
  • #1
koab1mjr
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Homework Statement


Let Sn be a convergent sequence and the limit of Sn > a. Prove that there exists a number N such that when n > N implies Sn > a


Homework Equations



sequence definition of limit

The Attempt at a Solution



This one does not seem to have a point to me. By definition when n > N the sequence is epsilon close to a and they tell us the limit is greater than a so once n>N we are close to the lim of Sn so we must be above a. Am I missing something?
 
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  • #2
Say lim Sn=b. And b>a. Just use the distance from b to a to make a particular choice of an epsilon.
 
  • #3
I would let epsilon be less than distance between a and b where b = lim Sn. With that selected as my epsilon can I use the fact that Sn being convergen means it can be slower than d(a,b) therefore that N exists?
 
  • #4
Well, yes. Pick e.g. epsilon=|b-a|/2. Now write out the proof.
 
  • #5
I think I got it now, my one question is why can't i just say epsilon less than d(a,b). That way when I say Sn is convergent b - epsilon is still bigger than a. You have d(a,b)/2. I thnk for the same reason but i might be overlooking something. Thanks for your help.
 
  • #6
No reason why you couldn't. If b= lim Sn> a, and [itex]\epsilon= d(a,b)= b-a[/itex] then there exist N such that if n>N then |Sn-a|< \epsilon= b-a. Then -(b-a)= a- b< b- Sn< b-a so, subtracting b from each part a- 2b< -Sn< -a. Multiplying each part by -1, 2b-a> Sn> a.
 

FAQ: Is There an N Where n > N Implies Sn > a in a Convergent Sequence?

1. What is "Another Sequence proof too trivial to figure out"?

"Another Sequence proof too trivial to figure out" refers to a mathematical proof involving a sequence that is considered too simple or obvious to solve, but has yet to be proven.

2. Why is this proof considered trivial?

This proof is considered trivial because it involves a sequence that is easy to understand and follow, making it appear simplistic and obvious to solve. However, finding a formal proof for it may still require advanced mathematical techniques.

3. Has anyone been able to figure out this proof?

As of now, the proof is still considered unsolved and remains a topic of interest among mathematicians. Some may have attempted to solve it, but no definitive answer has been found yet.

4. What is the significance of solving this proof?

Solving this proof would not only provide a better understanding of the underlying mathematical concepts, but it could also potentially lead to further advancements in the field of mathematics.

5. Is there any progress being made towards solving this proof?

There may be ongoing efforts by mathematicians to solve this proof, but due to its simplicity, it may not be receiving as much attention as more complex problems. However, with advancements in technology and new approaches to problem-solving, there is always a possibility of finding a solution in the future.

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