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Partial sum problem

  1. Nov 3, 2005 #1
    If the partial sum of a series is given by
    [itex] s_N= \tfrac 1{N} \cos(N \pi) [/itex]
    is it then possible to concl. that the series is convergent because [itex] s_N \rightarrow S =0 [/itex]
    if so, can one proove abs. convergence by noticing the same for [itex] s_N=\tfrac 1{N}[/itex] ?
    Thanks
     
  2. jcsd
  3. Nov 3, 2005 #2

    StatusX

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    Yes, that's true. If the sum of the first N terms in the series can be explicitly calculated, and you show what this converges to as N goes to infinity, you are summing the series the only way possible. Note the nth term in the series is S(n)-S(n-1). I'm not sure if you're asking why a series with partial sums 1/N converges while the series with terms 1/N diverges, but if you are, remember these are different series (the former having terms 1/N-1/(N-1)=-1/(N^2-N).)
     
  4. Nov 3, 2005 #3
    my main problem was to check for abs. convergence.
    bacause usually this is done by checking [itex] \sum |a_n| = |a_1| + |a_2| + ...[/itex], and i dont find it intuitive that the same result is reached by checking [itex] |a_1+a_2+...+a_N|=|s_N| \rightarrow S [/itex] as [itex] n \rightarrow \infty [/itex] ..
     
  5. Nov 3, 2005 #4
    This is absolutely not true. If a series converges, then the sequence {Sn}->0, but the converse is not true. Look at 1/N.
     
  6. Nov 3, 2005 #5
    the sequence {Sn} that they are talking about is the sequence of partial sums of the series in question. When the sequence of partial sums converges then you can conclude that the series converges.
     
  7. Nov 3, 2005 #6
    No, you cannot conclude that it is absolutely convergent because Sn approaches zero. That test only rules out convergence if the limit approaches anything BUT zero.

    As the series stands, it is an alternating harmonic series, and as such does converge, because.. 1. lim n approaches inf Sn = 0, and 2. For all large Sn, Sn+1 is less than Sn.

    But it does not converge, absolutely. In this sense, it is a run of the mill, harmonic series, which does NOT converge. You are correct in this comparison.. the series actually IS 1/N- and fails to converge, both by the integral and "p" test.
     
    Last edited: Nov 3, 2005
  8. Nov 4, 2005 #7
    given is two series [itex] \sum_{n=1}^\infty a_n [/itex] and [itex] \sum_{n=1}^\infty b_n [/itex] where [itex] b_n=|a_n| [/itex]. now if [itex] S_N= \frac 1{N} \cos (N \pi)[/itex] is the partial sum of [itex] \sum_{n=1}^\infty a_n [/itex] then this series converges because it is alternating and the sequence [itex] \{ a_n \} _{n=1}^\infty [/itex] decreases to zero (because [itex] S_N \rightarrow 0 [/itex]).
    Do you mean that [itex] b_n = \frac 1{n} [/itex]? Why?
     
  9. Nov 4, 2005 #8
    yes, Bn=1/n, if you take the absolute value of each term An.

    Cos (N PI) = (-1)^(N).

    So this makes An an alternating harmonic series, and it converges.

    But does it converge, absolutely? No. Take the absolute value of each term of An, then Bn is just a plain, non- alternating harmonic series.
     
    Last edited: Nov 4, 2005
  10. Nov 4, 2005 #9
    [itex] a_n [/itex] is not the partial sum, but the term in the series [itex] \sum_{n=1}^\infty a_n [/itex].
    [itex] S_N = a_1+a_2+...+a_N [/itex] is the partial sum. Surely the unknown series with partial sum [itex] S_N [/itex] cannot be [itex] \sum_{n=1}^\infty \tfrac 1{n} \cos (n \pi) [/itex] as you suggest, because i.e.
    [itex] \sum_{n=1}^5 \tfrac 1{n} \cos (n \pi) [/itex] never equals [itex] S_5 = a_1+...+a_5 = \tfrac 1{5} \cos (5 \pi) [/itex]
     
    Last edited: Nov 4, 2005
  11. Nov 4, 2005 #10
    Drats.. I edited that thing to say term.. oh well.

    No, what I am saying is..if I understand the notation correctly, the sum of the terms of the series (1/N)cos(N Pi) = (1)cos (1*Pi) + (1/2)cos(2*Pi) (1/3)cos (3*Pi) + (1/4)cos (4*pi) +(1/5)cos (5*Pi)..

    equals (1)(-1) + (1/2)(1) + (1/3)(-1) + (1/4)(1) + (1/5)(-1)...

    which equals -1 +1/2 -1/3 +1/4 -1/5... which makes it an alternating harmonic series. To prove convergence, all that I need is Lim n approaches inf An = zero.. as for all An, An+1 < An.

    The series does NOT sum to zero, if I understand it correctly..

    My impression is the answer to the first question, is yes.. the series is convergent.

    But.. is it absolutely convergent? No.
     
    Last edited: Nov 4, 2005
  12. Nov 4, 2005 #11
    This is not what is meant by ´partial sum´, rather we know that the N'th partial sum equals [itex] \sum_{n=1}^N = a_1+...+a_N = S_N [/itex] This does not mean that the terms in the series is (1/N)cos(N Pi) = (1)cos (1*Pi) + (1/2)cos(2*Pi) + (1/3)cos (3*Pi) + (1/4)cos (4*pi) +(1/5)cos (5*Pi) as you claim.

    This is also the argument i would use: Because [itex] S_N \rightarrow 0 [/itex] the series [itex] \sum_{n=1}^\infty a_n [/itex] is convergent with sum zero. But how can one check if the convergence is conditional or absolute?
     
  13. Nov 4, 2005 #12

    NateTG

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    What would you normally check to see if a series has conditional or absolute convergence?
     
  14. Nov 4, 2005 #13

    StatusX

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    I don't see what could be meant by the sum of an infinite series other than the limit of the partial sums. If this limit is finite, the series converges.
     
  15. Nov 4, 2005 #14
    I would check the series [itex] \sum_{n=1}^\infty | a_n| [/itex], wich is unknown...I dont see, what knowing the N'th partial sum of [itex] \sum_{n=1}^\infty a_n [/itex] tells me about [itex] \sum_{n=1}^\infty | a_n| [/itex]...help : ) surely its not as simple as saying that the N'te partial sum of [itex] \sum_{n=1}^\infty | a_n| [/itex] is |S_N| ?
     
    Last edited: Nov 5, 2005
  16. Nov 5, 2005 #15
    after thinking about it, i came to conclude that [itex] \sum_{n=1}^\infty a_n [/itex] is convergent, but not abs. convergent. I realized that i.e.
    [tex] \sum_{n=1}^N a_n = \sum_{n=1}^N (-1)^{n}(2n-1) \frac{|n-2|!}{n!}=\frac 1{N} \cos (N
    \pi) \mbox{ for all } N \in \mathbb{N} [/tex] and checked this series for convergence (used Maple), wich lead to the result: conditional convergence...
    but isent there a easier proof?
     
    Last edited: Nov 5, 2005
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