From the fact that you use the definition of sequence convergence.
But in place of just you are free to use .
Problem states:
Let be a sequence in with = a.
If a > 0, prove (the set of Natural numbers) such that .
My work so far:
I don't have a proof written up for this problem yet, but I have jotted down some ideas that I think might help me:
I know that the sequence { } converges to a and a > 0.
So by definition of convergence, I know that there exists an in the Naturals such that
| - a| < ε for all n ≥ .
There is also the theorem that every convergent sequence is bounded, so by definition of bounded: there exists an M > 0 such that
| | ≤ M for all n in the Naturals.
So for this proof do I need cases? Would I do proof by contradiction?
I worked with a study group on this, and my classmate drew a picture and talked us through the thinking. It kind of made sense to me that if a sequence converges to zero, then eventually the sequence has to be greater than zero at some point.
Any help, suggestions, and/or tips are greatly appreciated.
Thank you for your time! =)
You could do this proof by contradiction. Assume that there does not exist a such that
By definition of we must have that for every there exists a corresponding such that . So if we take we must be able to find a such that , but now by assumption which gives a contradiction.
Yes you are right, that is unclear. Let me fix it up.
Suppositions:
1.
2.
3. There exists a such that
A fact we will use, if and then
Now by the definition of a sequence converging to we must have that for every there exists a corresponding such that . So now let , so we must be able to find a such that . So it must be true that if we take that . But here we arrive at a contradiction. The reason being that because of supposition three we must have that