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Convergence of series (Theoretical Question)

  1. Jun 20, 2016 #1
    1. The problem:

    Ive been all afternoon struggling with this doubt. Its a bit more teoric than the rest of the exercices i did and i just cant seem to get around it so here it goes :
    --------------------------------------------------------------------------------------------------------------------------------

    Consider ∑ 1/an a convergent serie of positive terms.

    --------------------------------------------------------------------------------------------------------------------------------

    What´s the nature of Σ (-1)^n/(an^(2) +1)

    2. Relevant equations
    (Leibniz criterion) ? If ∑(-1)^(n)an is an alternated serie and the sucession(an) is decreasing and his limit→+00 an = 0 then i can say that ∑(-1)^(n)an is converging.

    3. The attempt at a solution


    My initial thought was:... a-ha! This is an alternate serie! Because of the (-1)^n ...

    So if i prove that the serie 1/an is decreasing since i already know the lim an= 0 i can say that the serie:
    (Leibniz criterion)


    Σ (-1)^n/(an) IS CONVERGENT however i have two problems...


    1- I dont know how to prove that 1/an is decreasing.

    2- The serie that they ask me to study is different and even if i could prove that 1/an is decreasing i dont know if through algebric manipulation i could get to:

    Σ (-1)^n/(an^(2) +1)




    Im not even sure if im going the right way. Anyone has any clue? This is bothering me so much!atement, all variables and given/known data
     
    Last edited by a moderator: Jun 20, 2016
  2. jcsd
  3. Jun 20, 2016 #2

    Mark44

    Staff: Mentor

    Given that ##\sum \frac 1 {a_n}## is a convergent series with ##a_n > 0##, can you say something about the series ##\sum \frac 1 {a_n^2 + 1}##?
    Are the terms of this series larger, smaller, or the same as, term by term, of the series ##\sum \frac 1 {a_n}##?
     
  4. Jun 20, 2016 #3
    The terms start in n= 1 and go up to +00
     
  5. Jun 20, 2016 #4

    Mark44

    Staff: Mentor

    Fine. Did you read my post?
     
  6. Jun 21, 2016 #5
    Oh sorry. yes i did, i notice that the terms are smaller. Are you saying that i can compare the serie
    Σ 1/(an^(2) +1) with (1/an) using the criterion of geral comparison?

    If an≤bn for n ≥ f then:

    (1) -Σbn is convergent → Σan is convergent

    (2) -Σan is divergent → Σbn is divergent

    At a first glance it doesnt seem that useful cause since the terms are smaller i get that Σ 1/(an^(2) +1) is (an) and that 1/an is (bn).

    So i guess now i can say that Σ 1/(an^(2) +1) is convergent?

    However that serie that im trying to prove doesnt have just positive terms, Σ (-1)^n/(an^(2) +1) also has negative terms so i cant apply criterion of comparison. So i have no clue...
     
    Last edited: Jun 21, 2016
  7. Jun 21, 2016 #6

    Mark44

    Staff: Mentor

    Yes.
    If you know that a series ##\sum b_n## converges, can you conclude anything about the alternating series ##\sum (-1)^nb_n##? I.e., does it
    a) converge,
    b) diverge, or
    c) is it impossible to determine?
     
  8. Jun 21, 2016 #7
    I cant conclude nothing using the geral criterion of comparison cause it only applies to series with positive terms. The problem is that (-1)^n !!!

    Im almost sure it converges but on the definition of this criterion it says it is explicitly for series with positive terms.
    ----------------------------------------------------------------------------------------------------
    I have this example in my head: Σ(1/n)

    1/2,1/3,1/4,1/5 1/6...

    If i join that (-1)^n it will be something like:

    1/2,-1/3,1/4,-1/5 1/6... which also converges but using a example obviously in no valid explanation.
    -------------------------------------------------------------------------------------------------------------------

    I also noticed that 1/an looks like a dirichlet series however i cant also make that assumption! cause the "(an)" can be anything!!! If i could im almost sure i knew how to do it.

    -----------------------------------------------------------------------------------------------------------------------
    is it valid to do something like this?

    1) Σ(-1)^n/(an^(2) +1) =( -1)^nΣ(1/(an^(2) +1)

    2) since i know 1/an is a serie of positives terms larger than Σ(1/(an^(2) +1) and 1/an is convergent , Σ(1/(an^(2) +1) is also convergent.

    3)Since Σ(1/(an^(2) +1) is convergent, ( -1)^nΣ(1/(an^(2) +1) also is (just cause? ).
     
    Last edited: Jun 21, 2016
  9. Jun 21, 2016 #8

    George Jones

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    At the risk of gving too much away,
     
  10. Jun 21, 2016 #9
    I know that if (bn) converges, (an) will also converge. So i know that Σ(1/(an^(2) +1) is convergent, However my problem is that the serie in question is Σ((-1)^n/(an^(2) +1) and since it has negative terms i cant apply that criterion. The same applies with bn. If i know bn is convergent what criterion should i use to consider bn(-1)^n also convergent?

    What is on my mind is: If i know that a certain serie is convergent, the limit →+00 = 0 and it doesnt seem to change if just multiply it by (-1)^n
     
  11. Jun 21, 2016 #10

    George Jones

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    Forget everything except that there is a series with terms ##b_n##, and that the series ##\sum b_n## converges. What can you conclude?
     
  12. Jun 21, 2016 #11
    I can conclude that bn limit →+00 = 0 and that an is convergent.
     
  13. Jun 21, 2016 #12

    Ray Vickson

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    Go right back to basics. If ##b_n > 0## and ##S =\sum b_n## is convergent. By definition, this means that
    [tex] S(N) = \sum_{n=1}^N b_n [/tex]
    is a convergent sequence; that is, ##S = \lim_{N \to \infty} S(N) ## exists and is finite. Note that
    [tex] \sum_{n=1}^N b_n = \sum_{n \leq N \;\text{even}} b_n + \sum_{n \leq N \;\text{odd}} b_n . [/tex]
     
  14. Jun 21, 2016 #13
    Ok! i think i got it. Can anyone just say if this sounds like a reasonable answer?

    They ask me to study nature of ∑(-1)^n/(an^(2) + 1) knowing that 1/an is convergent.

    Step 1- I know that ∑(1/(an^(2) + 1) < 1/an .

    Step 2- Since all the terms of the series in Step 1 are positive i can conclude by the criterion of geral comparison that since 1/an is convergent ∑(1/(an^(2) + 1) is also convergent.

    Step 3- Consider bn = ∑(1/(an^(2) + 1) so
    ∑bn*(-1)^n = ∑(-1)^n/(an^(2) + 1)

    According to leibniz criterion, if ∑ bn*(-1)^n is an alternate serie where the sucession bn is decreasing and convergent and its lim+00 = 0.(basically saying that bn must be convergent)
    I can conclude that ∑ bn*(-1)^n = ∑(-1)^n/(an^(2) + 1) is convergent!
     
    Last edited: Jun 21, 2016
  15. Jun 21, 2016 #14

    Ray Vickson

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    I can come up with a sequence in which the ##b_n>0 ## are not monotone non-increasing, but which go to zero quickly enough that the series ##\sum b_n## converges anyway. You could not apply your argument to such a case.
     
  16. Jun 21, 2016 #15

    Ray Vickson

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    Your Step 1 is wrong; you may have ##1/(a_n^2 +1) < 1/a_n## but there is no reason at all to assume you have ##\sum 1/(a_n^2 +1) < 1/a_n##. When you write things like that you are just begging for marks to be taken off!

    Anyway, as I said before, you cannot just apply an alternating-series argument, because the successive terms may not be decreasing each and every time.
     
  17. Jun 21, 2016 #16
    Sorry, ive been reviewing the steps and what i really meant to write as Step 1 was :


    Step 1- I know that (1/(an^(2) + 1) < 1/an . (i bugged the integral symbol on the previous post)

    Step 2- Since Σ1/an and ∑(1/(an^(2) + 1) have just positive terms i can conclude by the criterion of geral comparison that since ∑1/an is convergent ∑(1/(an^(2) + 1) is also convergent.

    Step 3- Consider bn = ∑(1/(an^(2) + 1) so
    ∑bn*(-1)^n = ∑(-1)^n/(an^(2) + 1)

    According to leibniz criterion, if ∑ bn*(-1)^n is an alternate serie where the sucession bn is decreasing and convergent and its lim+00 = 0.(basically saying that bn must be convergent)
    I can conclude that ∑ bn*(-1)^n = ∑(-1)^n/(an^(2) + 1) is convergent!


    Note:Since bn is convergent cant i tell by D`Alembert criterion that if i do lim n->+00 bn+1/bn = <1 and so... doesnt this mean that bn is decreasing?
     
    Last edited: Jun 21, 2016
  18. Jun 21, 2016 #17

    Ray Vickson

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    No. Consider an example like ##b_n = |\sin(n)|/n^{1.1}##. We have ##b_n \leq 1/n^{1.1}##, so ##\sum b_n## is dominated by the convergent ##p##-series ##\sum 1/n^{1.1}##, but the terms of ##b_n## do not form a non-decreasing sequence: we can have ##b_{n+1} > b_n## for infinitely many values of ##n##. Nevertheless, we can establish the convergence of ##\sum (-1)^n b_n##, essentially using the hints I supplied in post #12.
     
  19. Jun 22, 2016 #18
    I´ve read and spent some time with post 12, but i just dont see how it does prove the convergence of ∑(-1)^nbn, could you provide me with any hint?
     
  20. Jun 23, 2016 #19

    Ray Vickson

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    Try writing out the first six or seven terms of both the finite sums ##\sum_{n=1}^N b_n## and ##\sum_{n=1}^N (-1)^n b_n##, then look carefully at what you have. It is important to remember that for finite sums it is perfectly legitimate to re-order the terms and collect them in any convenient forms you want.
     
    Last edited: Jun 23, 2016
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