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Analysis - Cauchy sequences?

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove the following theorem, originally due to Cauchy. Suppose that [itex](a_{n})[/itex][itex]\rightarrow a[/itex]. Then the sequence [itex](b_{n})[/itex] defined by [itex]b_{n}=\frac{(a_{1}+a_{2}+...+a_{n})}{n}[/itex] is convergent and [itex](b_{n})[/itex][itex]\rightarrow a[/itex].


    2. Relevant equations
    A sequence [itex](a_{n})[/itex] has the Cauchy property if, for each [itex]ε>0[/itex] there exists a natural number N such that [itex]|a_{n} - a_{m}|<ε[/itex] for all [itex]n,m>N[/itex]


    3. The attempt at a solution
    I don't exactly know what am I suppose to do here ... ?
    The only thing I could think of is [itex](a_{1}+a_{2}+...+a_{n})≈n*a_{n}[/itex] for a very huge n but that doesn't seem really relevant.

    Am I suppose to show that [itex](b_{n})[/itex] is a subsequence? Or [itex](b_{n})[/itex] is somewhat related to the [itex](a_{m})[/itex] in the definition of the Cauchy sequence?
     
    Last edited: Nov 7, 2012
  2. jcsd
  3. Nov 7, 2012 #2

    LCKurtz

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    You know ##a_n\rightarrow a## and you want to make$$
    \left| \frac{a_1+a_2+...+a_n}{n}-a\right |$$small if ##n## is large enough. Start with that.
     
  4. Nov 7, 2012 #3
    Case 1: [itex](a_{n})[/itex] is an increasing sequence
    [itex]\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a<\frac{n*a}{n}-a=a-a=0
    \Rightarrow b_{n}-a<0[/itex]

    Case 2: [itex](a_{n})[/itex] is a decreasing sequence
    [itex]\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a>\frac{n*a}{n}-a=a-a=0
    \Rightarrow b_{n}-a>0[/itex]

    [itex]\Rightarrow 0<b_{n}-a<0

    \Rightarrow |b_{n}-a|<0

    \Rightarrow b_{n}\rightarrow a[/itex]

    Does this make any sense?
     
  5. Nov 7, 2012 #4

    LCKurtz

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    I'm afraid not. The sequence can't be both increasing and decreasing unless it is a constant, and it may not be either one anyway. I would try to find an ##\epsilon,\, N## type of argument.
     
  6. Nov 7, 2012 #5

    LCKurtz

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    I have to leave for a couple of hours. Let me just add not to waste your time using the Cauchy criteria. All you need is the ##\epsilon,\, N## definition of ##a_n\rightarrow a##.
     
  7. Nov 7, 2012 #6
    Okay, thanks! Will try ...
     
  8. Nov 8, 2012 #7
    [itex]a_{n}\rightarrow a \Rightarrow |a_{n}-a|<\epsilon, \forall n>N[/itex]

    Assume [itex]|\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a|<2\epsilon[/itex]

    [itex]|\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a-(a_{n}-a)| \leq |\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a|-|a_{n}-a| < 2\epsilon-\epsilon[/itex]
    [itex]\Rightarrow |\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a_{n}|<\epsilon[/itex]

    Okay I honestly have no idea where am I going with this. :cry:
     
  9. Nov 8, 2012 #8

    LCKurtz

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    Given ##\epsilon > 0## there exists ##N## such that
    If you are going to do an ##\epsilon,\, N## proof, it is best to give complete and exact statements. And it might be a good idea to use ##\epsilon/2## or ##\epsilon/4## to give yourself some extra room.
    But that is basically what you are trying to prove. You can't start by assuming it is true.

    Yes, I see that. As I said earlier, you want to start with $$
    \left| \frac{a_1+a_2+...+a_n}{n}-a\right |$$
    and make it small. To get started write it as a single fraction$$
    \left| \frac{a_1+a_2+...+a_n-na}{n}\right |$$Now since, at least for large ##n##, you can make ##|a_n-a|## small, try spreading the n a's with each of the ##a_i##'s. Then think about the part where ##n\le N## and where ##n>N## and see if you can make them both small for ##n## large enough.
     
  10. Nov 8, 2012 #9
    Hmm ... guess I should refrain from rephrasing definitions so carelessly.

    ##|b_{n}-a|=|\frac{(a_{1}+a_{2}+...+a_{n})}{n}-a|##
    ##=|\frac{(a_{1}+a_{2}+...+a_{n})-na}{n}|##
    ##=|\frac{((a_{1}-a)+(a_{2}-a)+...+(a_{n}-a))}{n}|##
    ##\leq |\frac{(a_1-a)}{n}|+|\frac{(a_2-a)}{n}|+...+|\frac{(a_n-a)}{n}|## [by Triangle Inequality]
    [itex]<\frac{\epsilon_1}{n}+\frac{\epsilon_2}{n}+...+[/itex][itex] \frac{\epsilon_{n}}{n}[/itex] [Since ##|a_n-a|<\epsilon## and the theorem that states that subsequences (i.e. ##a_1, a_2, ...##) tend to the same limit (single term subsequences as well??)]
    ##=\frac{n\epsilon}{n}##
    ##=\epsilon##
    [itex]\Rightarrow |b_{n}-a|<\epsilon[/itex]

    Obviously missing out on something (or totally off track?) since there was nothing relevant to your last sentence. I understand that by definition of convergent sequences, it will only be between both ##\epsilon## if ##n## is larger than a specific point ##N## that corresponds to the ##\epsilon## chosen but ... then what?
    I feel very bad for having the lack of capability to comprehend your hints ...
     
  11. Nov 8, 2012 #10

    LCKurtz

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    OK so far. Just do what I suggested. If ##n>N## part of that sum has ##n\le N## and part has ##n>N##. Break it into those two parts. Then what you have to figure out is why both parts are small if ##n## is large enough (and how large "large enough" is). That is the meat of the argument.
     
  12. Nov 8, 2012 #11
    I guess for ##n\le N##, this part would be small enough because of a large denominator, i.e. ##n##? Whereas for ##n>N, (a_n-a)## would be small ... but I'm pretty sure this isn't what you're alluding to ...

    I don't get it :confused:

    What end result am I looking for? Something like the one I posted before but with different intermediate steps?
     
  13. Nov 8, 2012 #12

    LCKurtz

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    That is the idea but you must write it mathematically. And you haven't yet broken it into two parts depending on ##N##. You are going to need an ##N## in there do do that. Then show how big and why that big that ##n## must be to make each of those less than ##\epsilon/2## to get the whole thing less than ##\epsilon##.
    What are you looking for?? If you don't know that no wonder you are confused. Read your statement in red above. And ##b_n\rightarrow a## means given ##\epsilon > 0## blah blah. You really need to learn carefully to write down what you are given and what you are trying to prove, using the definitions.
     
  14. Nov 8, 2012 #13
    Something like this??
    ##|\frac{(a_1-a)}{n}|+|\frac{(a_2-a)}{n}|+...+|\frac{(a_N-a)}{n}|+|\frac{(a_{N+1}-a)}{n}|+...+|\frac{(a_n-a)}{n}|##
    ##(|\frac{(a_1-a)}{n}|+...+|\frac{(a_N-a)}{n}|)+(|\frac{(a_{N+1}-a)}{n}|+...+|\frac{(a_n-a)}{n}|)##

    To clarify, I wasn't unsure of the main objective of the question, just wanted to know whether am I suppose to get something something ##<\epsilon##, to which you have answered.
     
  15. Nov 8, 2012 #14

    LCKurtz

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    That's better. Now look again at the advice from post #6 I have highlighted above.
     
  16. Nov 9, 2012 #15
    Only thing I could come up with is this:
    ##(|\frac{(a_1-a)}{n}|+...+|\frac{(a_N-a)}{n}|)<\frac{|a_n-a|}{2}<\frac{\epsilon}{2}##
    ##\frac{N(a_N-a)}{n}<\frac{|a_n-a|}{2}##
    ##\frac{N}{n}<\frac{1}{2}##
    ##n>2N##
     
  17. Nov 9, 2012 #16
    Okay, thought about it again (okay not really, kinda discussed with my friends) and this is the result.

    Let ##\epsilon>0, |a_n-a|<\frac{\epsilon}{2} \ \forall n>N##
    ##|\frac{a_{N+1}-a}{n}|+...+|\frac{a_{n}-a}{n}|<\frac{\epsilon_{N+1}}{2n}+...+\frac{ε_{n}}{2n}=(\frac{n-N}{n})(\frac{\epsilon}{2})##
    Since ##n>N, (\frac{n-N}{n})<1##
    ##\Rightarrow|\frac{a_{N+1}-a}{n}|+...+|\frac{a_{n}-a}{n}|<\frac{\epsilon}{2}##

    For the remaining part ...
    ##|\frac{a_{1}-a}{n}|+...+|\frac{a_{N}-a}{n}|=\frac{1}{n}(|a_1-a|+...+|a_N-a|)##
    ##(|a_1-a|+...+|a_N-a|)## is already fixed, therefore for a large enough ##n, \frac{1}{n}(|a_1-a|+...+|a_N-a|)<\frac{\epsilon}{2}##

    I still don't see how can I show how big of an ##n## is required ...
     
  18. Nov 9, 2012 #17

    LCKurtz

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    Solve that inequality for n to see how large it must be.
     
  19. Nov 9, 2012 #18
    ##\frac{1}{n}(|a_1-a|+...+|a_N-a|)<\frac{\epsilon}{2}## this?
    ##\frac{1}{n}(|a_1-a|+...+|a_N-a|)<\frac{\epsilon}{2}##
    ##\frac{2(|a_1-a|+...+|a_N-a|)}{\epsilon}<n##
     
  20. Nov 9, 2012 #19

    LCKurtz

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    I didn't notice when I scanned your post a minute ago. Where did all these ##\epsilon_n##'s come from? You can't introduce new variables in a proof without saying what they represent.
     
  21. Nov 9, 2012 #20
    Oh, they're just ##\epsilon##'s that correspond to each term ... which are essentially the usual ##\epsilon##'s. Sorry about that.
     
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