Limit proof on monotonic sequences

  • Thread starter Bipolarity
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  • #1
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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
 

Answers and Replies

  • #2
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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.
 
  • #3
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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.
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
 
  • #4
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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?
 
  • #5
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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?
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
 
  • #6
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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
 
  • #7
22,089
3,286
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
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.

Hey micromass, I have an idea.
What if we let [itex] ε = a_{n-1} [/itex] ?

BiP
Yeah ok. How would that work?
 
  • #8
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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
 
  • #9
22,089
3,286
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
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??
 
  • #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]?
 
  • #11
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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?
 
  • #12
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Ok, somehow the font got messed up, but just to make it clear, [a][/1] is meant to be a1.
 

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