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Math Help - Cauchy Sequence

  1. #1
    Senior Member sfspitfire23's Avatar
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    Cauchy Sequence

    Hey guys, I'm having trouble wrapping my head around Cauchy Sequences.

    So, I know that the definition of such a sequence is if  m,n>N then  |a_n-a_m|<\epsilon. This means that the sequence is getting epsilon close to each other far down the number line.

    How do we know what to choose for a_m and a_n if we say n>m>N when trying to prove that the sequence X_n=1/n is Cauchy?


    Also, the proof "if a sequence converges, then it is Cauchy," (bottom of first page here: http://legacy.lclark.edu/~istavrov/a...t30-cauchy.pdf)

    why is it true that |s_n-s|<\epsilon /2? Why is it less than epsilon over two and not just epsilon. I assume it has something to do with the triangle inequality?

    If theres any trick to understanding this stuff please share!


    Thanks!
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  2. #2
    Member
    Joined
    Feb 2010
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    Quote Originally Posted by sfspitfire23 View Post
    Hey guys, I'm having trouble wrapping my head around Cauchy Sequences.

    So, I know that the definition of such a sequence is if  m,n>N then  |a_n-a_m|<\epsilon. This means that the sequence is getting epsilon close to each other far down the number line.

    How do we know what to choose for a_m and a_n if we say n>m>N when trying to prove that the sequence X_n=1/n is Cauchy?


    Also, the proof "if a sequence converges, then it is Cauchy," (bottom of first page here: http://legacy.lclark.edu/~istavrov/advcalc-sept30-cauchy.pdf)

    why is it true that |s_n-s|<\epsilon /2? Why is it less than epsilon over two and not just epsilon. I assume it has something to do with the triangle inequality?

    If theres any trick to understanding this stuff please share!


    Thanks!

    I think you're making things a bit more complicated than they have to be.

    Lets start with "if a sequence converges, then it is Cauchy," and forget the formal proof for a second. Intuitively, what this says is that if a sequence is getting closer and closer to a single point, then the points in the sequence must be getting closer and closer to each other. Makes sense, right?

    As for the proof, since the sequence converges, we have that (sn) approaches s as n goes to infinity. Thus for any number (here, that number is epsilon/2), there is an N such that if n>N then |s_n-s|<\epsilon /2. This is because |s_n-s| approaches 0 as n-->infinity. The key point is that epsilon is arbitrarily chosen.


    As for your question, simply find an N such that 1/N < \epsilon /2....and try go from there.
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