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# prove if (a_n)--> infinity then (1/a_n) --> infinity watch

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1. (Original post by hiyatt)
is there a nicy easy fix to this without re doing the whole thing
The whole proof is about 3 lines! How easy do you want it to be?
2. (Original post by DFranklin)
The whole proof is about 3 lines! How easy do you want it to be?
lool you're funny. can i just let e=1/C at the end so it matches.
3. (Original post by DFranklin)
The thing is, it can be hard to distinguish between inability to put everything in a coherent logical order, and just a reluctance to actually post any mathematics here.

Or at least, posters frequently post multiple single line "so, should I do this...?" type questions, and you have no idea what their actual proof looks like anymore...
Very true!
4. (Original post by hiyatt)
lool you're funny. can i just let e=1/C at the end so it matches.
No. That's the 3rd time you've asked if you can do this. The answer isn't going to change.
5. (Original post by ftfy)
Very true!
That post of mine is looking staggeringly prophetic given the OPs most recent postings...
6. (Original post by DFranklin)
No. That's the 3rd time you've asked if you can do this. The answer isn't going to change.
I have said at the begginging to let C>0 so if you do that you see e varies to in the required range though
7. (Original post by hiyatt)
I have said at the begginging to let C>0 so if you do that you see e varies to in the required range though
That's not the way around it works. You have to *start* by choosing an arbitrary epsilon > 0. Once you do that, it's fixed - you can't set it to 1/C later.
8. (Original post by DFranklin)
That's not the way around it works. You have to *start* by choosing an arbitrary epsilon > 0. Once you do that, it's fixed - you can't set it to 1/C later.
9. (Original post by hiyatt)
You should think in particular about the question I asked you in post 17:
Now, you want to show |1/a_n| < epsilon (for sufficiently large n). What condition does this put on a_n?
10. (Original post by DFranklin)
That post of mine is looking staggeringly prophetic given the OPs most recent postings...
Indeed!
11. (Original post by DFranklin)
You should think in particular about the question I asked you in post 17:

can you do the work to get to C/2 and then say if we choose a e>0 there exists an N such that we have l1/a_nl<e whenever n>N
12. (Original post by hiyatt)
can you do the work to get to C/2 and then say if we choose a e>0 there exists an N such that we have l1/a_nl<e whenever n>N
If you think you have a proof, post a proof. It's impossible to judge whether a partially complete fragment would "work", although the fact that you talk about C/2 makes me pretty certain you've done something wrong.

13. (Original post by DFranklin)
If you think you have a proof, post a proof. It's impossible to judge whether a partially complete fragment would "work", although the fact that you talk about C/2 makes me pretty certain you've done something wrong.

Ok if this is not correct i will answer the question.

Prove 'If (a_n) --> infinity then (1/a_n) --> 0

As a_n--> infinity, for every C>0 there exists an N in the natural numbers such that we have

a_n>C whenever n>N... this implies

la_nl>C whenever n>N... this implies

l1/a_nl < 1/C whenever n>N

If follows that if we choose e>0 there exists an N_1 such that we have

l1/a_nl < e whenever n>N_1
14. (Original post by hiyatt)
Ok if this is not correct i will answer the question.

Prove 'If (a_n) --> infinity then (1/a_n) --> 0

As a_n--> infinity, for every C>0 there exists an N in the natural numbers such that we have

a_n>C whenever n>N... this implies

la_nl>C whenever n>N... this implies

l1/a_nl < 1/C whenever n>N

If follows that if we choose e>0 there exists an N_1 such that we have

l1/a_nl < e whenever n>N_1
It's still not right. A kind examiner might fill in the gaps, but he shouldn't have to. The line where you says "it follows" is particularly dubious.
15. (Original post by DFranklin)
It's still not right. A kind examiner might fill in the gaps, but he shouldn't have to. The line where you says "it follows" is particularly dubious.
can you write what you would say...
16. (Original post by hiyatt)
can you write what you would say...
No. In fact, I'm done with this. I can't be bothered repeating myself 5 times on every single thiing because you don't want to believe what I tell you.
17. (Original post by DFranklin)
No. In fact, I'm done with this. I can't be bothered repeating myself 5 times on every single thiing because you don't want to believe what I tell you.
Haha no of course i believe you you are much mor eintelligent than me i was just wondering if my version of it had any credibility

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Updated: January 10, 2019
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