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Is this a trivial proof?

  1. May 5, 2010 #1
    proof with the definition of the limit the following assertion: if a_n is a zero sequence, so is b_n with b_n = (1/n)* sum{a_j} [from j=1 to n].

    at first glance it seems trivial to me, since 1/n is a zero sequence so is b_n but apparently this doesnt satisfy the implication so how can i show that if a_n is a zero sequence, so b_n?
  2. jcsd
  3. May 5, 2010 #2
    seems so at first glance, let me think for any contradictions..
  4. May 5, 2010 #3
    By a zero sequence do you mean its like {0, 0, 0, 0, 0...} or it converges to 0 as [tex] n\rightarrow\infty[/tex]
  5. May 5, 2010 #4


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    Almost there...
    1/n is a zero sequence so too is C/n for any constant C
    consider C such that ||a_n||<C for all n (where ||a_n|| is the size of a_n
    Last edited: May 5, 2010
  6. May 5, 2010 #5
    Even though [itex]\{a_{n}\}[/itex] might be a zero sequence, the sequence of partial sums:

    S_{n} = \sum_{j = 1}^{n} {a_{j}}

    might not be convergent. A typical example is the harmonic series, which is divergent. So, if [itex]\{b_{n}\}[/itex] is a zero sequence and [itex]S_{n}[/itex] is divergent, you will get an indefinite form [itex]0 \cdot \infty[/itex] and it is not that trivial to show that this is a finite number.
  7. May 5, 2010 #6


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    That is needlessly complicated. It is trivial to show. S_n (which need not even be considered) may diverge, but it diverges like 1 thus S_n/n converges. It is like saying we need to consider an indeterminant form to decide if {1,1,...} converges since 1=n/n and n diverges, there is no such need.
    Last edited: May 5, 2010
  8. May 5, 2010 #7
    Actually, you need to prove that [itex]S_{n} = o(n), n \rightarrow \infty[/itex], so that [itex]b_{n}[/itex] would be a zero sequence.
    Last edited: May 5, 2010
  9. May 6, 2010 #8
    1/n has limit 0, so per definition has b_n ( 0*the sum).
    but the question is an implication, if a_n then b_n.

    |a_n - 0| <= delta for all n >= N (limit 0, zero sequence)

    the question is, is the sum of all the terms of the sequence a zero sequence, therefore limit 0.
  10. May 6, 2010 #9
    (a is the limit):

    choose delta so that |a_n - a| <= delta/n

    so for the last two terms of the sum:

    a_n+a_n-1 <= |a_n + a_n-1| =|a-(a-a_n)+a_n-1| <= |a_n -a| + |a_n-1 - a| < 2*delta/n (a=0 for the limit)

    and this applies for the rest of the sum a_j so sum{a_j} [from j=1 to n] < delta
    so far its convergence.
    is this argumentation valid?
    anyway, how can i show that its limit is zero (with the factor 1/n)?
  11. May 6, 2010 #10


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    suppose (as we know since {a_n}->0)
    for all epsilon>0 there exist an integer N (which may depend onepsilon) such that for all n>N
    Last edited: May 6, 2010
  12. May 6, 2010 #11
    is this equivalent with the decleration b_n=(1/n)(a_1+a_2+...+a_n) ?
    anyway i am not able to follow that, its likely a trivial fact which i cannot see, but someone could explain, please.
  13. May 6, 2010 #12


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    Yes it should have been the same, but for my error. The idea is all the a_k past some point are small, so the arithmatic average can be made small though a combination of making some elements small and making the nonsmall elements few.
  14. May 8, 2010 #13
    IMO "zero sequence" should mean [itex]0,0,0,\ldots[/itex] only, but I guess it means a sequence [itex]a_1,a_2,a_3,\ldots[/itex] such that [itex]a_n\to 0[/itex] now...

    Are you guys sure that that's standard terminology?
  15. May 8, 2010 #14
    You cannot justify a claim being trivial by not finding a counter example! :tongue: You mean the claim seems true, but you are thinking about counter examples just in case there might be some?

    This answers the original question. The correct answer is that the claim is not trivial.

    But it seems that the actual claim is true anyway... even though it's not trivial.
  16. May 9, 2010 #15
    well, the original question was to show the implication if a_0 then b_0.
    although b_0 is a zero sequence because of the zero sequence factor (1/n), one has to show that the sum of a_k is bounded so that b_n does not grow arbitrary. the trick is to rearrange the sum properly but i didnt manage that sofar.
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