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Limit proof on monotonic sequences

  1. Apr 29, 2012 #1
    Consider the sequence [itex] \{ a_{n} \} [/itex]

    If [tex] |a_{n+1}| > |a_{n}| [/tex]

    Prove that

    [tex] \lim_{n→∞} a_{n} ≠ 0 [/tex]

    The problem is part of a proof I am trying to understand, but I don't understand this particular step in the proof. Any ideas on how I might grasp this step?

    BiP
     
  2. jcsd
  3. Apr 29, 2012 #2

    micromass

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    Do you see it intuitively?? Did you draw a real number line and draw what such a sequence looks like??

    Of course I can just give you the proof, but I prefer that you "see it". If you do, then I believe the proof should be very easy.
     
  4. Apr 29, 2012 #3
    So because the sequence increases forever, it can't be 0 at infinity because if it were, then the term after that would be positive contradicting the fact that 0 is the limit?

    But how would I prove this using the ε-δ definition?

    BiP
     
  5. Apr 29, 2012 #4

    micromass

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    Yeah, so basically, the closest the sequences ever gets to 0 is in [itex]|a_0|[/itex]. After that it gets further and further away.

    So, what is the definition of [itex]\lim_{n\rightarrow +\infty}{a_n}=0[/itex]?? Can you find an ε that works?
     
  6. Apr 29, 2012 #5
    For all positive ε, we can find some N such that whenever n>N, then [itex] |a_{n}|<ε [/itex]. I am working on an appropriate value for ε as we speak.

    BiP
     
  7. Apr 29, 2012 #6
    Hey micromass, I have an idea.
    What if we let [itex] ε = |a_{n-1}| [/itex] ?

    Then that means there exists N>0 such that whenever n>N, then it must be the case that
    [itex] |a_{n}| < |a_{n-1}| [/itex]

    But am I allowed to shift the index of the term by 1 so I can show that it contradicts the premise?

    BiP
     
  8. Apr 29, 2012 #7

    micromass

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    Yeah, that is the definition for [itex]\lim_{n\rightarrow +\infty} a_n=0[/itex]. But you want to prove that this is NOT true. So you want the negation.

    Yeah ok. How would that work?
     
  9. Apr 29, 2012 #8
    Hey micro I just edited my last post. I think it is right, but I confused on why we are allowed to shift the index by one.

    In other words, so far we have shown that if [tex] \lim_{n→∞} a_{n} = 0 [/tex] then it must be the case that for some N>0 whenever n>N, then [tex] |a_{n}|<|a_{n-1}| [/tex] which says that the sequence is decreasing.

    But the original constraint of the problem is [tex] |a_{n}|<|a_{n+1}| [/tex].

    I must have made mistake somewhere or not chosen by epsilon correctly.

    BiP
     
  10. Apr 30, 2012 #9

    micromass

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    Yeah sure you can shift the index. You are given that [itex]|a_{k+1}|>|a_k|[/itex]for ALL k. So just choose k=n-1.

    But something bothers me. You have not defined n. You can't work with [itex] \varepsilon = |a_{n-1}|[/itex] because you didn't declare what n is.
    Why don't you pick a specific value of n??
     
  11. Apr 30, 2012 #10

    HallsofIvy

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    Choose an arbitrary n, as micromass suggests, and let [itex]\epsilon= |a_n|[/itex], for that specific n. What happens to the definition of [itex]\lim_{n\to\infty} a_n= 0[/itex] using that [itex]\epsilon[/itex]?
     
  12. Apr 30, 2012 #11
    Why don't we just pick [a][/1] as ε? Then |0-[a][/n]| < [a][/1] for all n > N, a contradiction. Or have I missed something?
     
  13. Apr 30, 2012 #12
    Ok, somehow the font got messed up, but just to make it clear, [a][/1] is meant to be a1.
     
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