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Cauchy Criterion for Series

  1. Nov 8, 2004 #1
    I know that if the series of (a)n (n is a subscript) converges, then the lim (a)n=0. How can I show that if the series of (a)n converges, then lim n(a)n=0?

    Or rather if a1 +a2 +a3 +...+an=0, then lim n*(a)n=0?

    Not sure how to show this, but I know the proof involves the cauchy criterion for series. Help anyone?
     
  2. jcsd
  3. Nov 8, 2004 #2

    Galileo

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    Hint:
    Let:
    [tex]S_n=\sum_{k=1}^na_k[/tex]

    Notice that
    [tex]a_n=S_n-S_{n-1}[/tex]
     
  4. Nov 8, 2004 #3
    I am sorry but can I ask you for another hint. I understand that what you wrote is true, but what am I supposed to do with it. Using the Caucy Criterion for series, I know that there is an N such that for all n>m>N,
    sn-sm< for all epsilon >0. But where do I go from here? Sorry for my slowness in comprehension.
     
    Last edited: Nov 8, 2004
  5. Nov 8, 2004 #4

    Galileo

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    Oh, actually my hint goes with a proof that doesn't use the Cauchy criterion.
    Just assume the series is convergent and take the limit on both sides of the equation.

    Ok, so the Cauchy criterion is:
    A series is a Cauchy-series if for every [itex]\epsilon>0[/itex] there is a N>0, such that [itex]|S_n-S_m|<\epsilon[/itex] voor any n,m>N.

    In particular, it holds for m=n+1.
    Now write out [itex]|S_n-S_m|[/itex] and see what you get for m=n+1.
     
  6. Nov 8, 2004 #5
    So this is sort of an inductive proof? We show the inequality holds for n=m+1, and since the an terms are non increasing, we know Sn-Sm<epsilon will hold where n>m+1. Correct?
     
  7. Nov 8, 2004 #6

    Galileo

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    NO! It's not an inductive proof.
    We have to show that IF a series [itex]\sum a_n[/itex] converges, then [itex]\lim_{n\rightarrow \infty} a_n= 0[/itex].

    So we ASSUME our series is convergent, then we know that for any [itex]\epsilon >0[/itex] we can find an N such that [itex]|Sn-Sm|<\epsilon[/itex] for all n,m>N.

    From the above assumption we have to show that:
    for any [itex]\epsilon>0[/itex] there exists an N, such that [itex]|a_n|<\epsilon[/itex] whenever n>N.
    (This is just the definition of [itex]\lim_{n\rightarrow \infty} a_n= 0[/itex].)

    What I meant was. If you plug m=n+1 in |Sm-Sn| and write it in terms of sums, then....

    (it's still a hint).
     
    Last edited: Nov 8, 2004
  8. Nov 9, 2004 #7

    HallsofIvy

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    No, that wasn't the original question.

    The problem was to prove that "If [itex]\Sigma a_n[/itex] converges, then
    [itex]lim_{n\rightarrow \infty}na_n= 0[/itex].
     
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