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Convergent -divergent series

  1. Nov 30, 2009 #1
    Suppose the series [tex]\sum a_{n}[/tex] diverges to [tex]+\infty[/tex],

    Then if the series does not diverge to infinity it means that the series converges, and

    consequently the statement : if [tex]\sum a_{n}[/tex] diverges ,then [tex]\sum b_{n}[/tex] diverges,is equivalent to :

    if [tex]\sum b_{n}[/tex] converges ,then [tex]\sum a_{n}[/tex] converges??
     
  2. jcsd
  3. Nov 30, 2009 #2
    You need an additional condition on the [itex]b_n[/itex]; if [itex]\sum a_{n}[/itex] diverges and [itex]|b_n| \geq |a_n|[/itex] for all n then [tex]\sum b_{n}[/tex] diverges also. & conversely, if [tex]\sum a_{n}[/tex] converges and [itex]|b_n| \leq |a_n|[/itex] for all n then [tex]\sum b_{n}[/tex] converges. (it's called the comparison test)

    ps- to anybody else who knows, how do I put the latex in line?
     
    Last edited: Nov 30, 2009
  4. Nov 30, 2009 #3
    You put it in using an [itex] tag.
     
  5. Dec 1, 2009 #4
    So if we say that :if [tex]\sum a_{n}[/tex] diverges, then prove that ,[tex]\sum(1+1/n)a_{n}[/tex] diverges ,

    it is not equivalent to :

    if [tex]\sum(1+1/n)a_{n}[/tex] converges ,then [tex]\sum a_{n}[/tex] converges
     
  6. Dec 1, 2009 #5
    Contrapositive?
     
  7. Dec 1, 2009 #6
    if [tex]b_{n} = (1+1/n)a_{n}[/tex] then they're equivalent. they're converses of each other
     
  8. Dec 1, 2009 #7
    Assuming the series [tex]\sum a_{n}[/tex] diverges to +[tex]\infty[/tex], then we have to show that the series [tex]\sum(1+1/n)a_{n}[/tex] diverges to +[tex]\infty[/tex].

    But according to contrapositive law it is equivalent to show:

    if [tex]\sum(1+1/n)a_{n}[/tex] does not diverge to [tex]+\infty[/tex] ,then [tex]\sum a_n}[/tex] does not diverge to [tex]+\infty[/tex]

    is that O.K
     
  9. Dec 1, 2009 #8
    that's right, although people usually just say converge rather than "does not diverge" even though they mean the same thing. & if this is for a homework problem you should probably mention the comparison test somewhere.
     
    Last edited: Dec 1, 2009
  10. Dec 2, 2009 #9

    HallsofIvy

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    The statement, "if the series does not diverge to infinity it means that the series converges" is true for positive series. The series [itex]\sum_{n=0}^\infty (-1)^n[/itex] neither diverges to infinity nor converges.
     
  11. Dec 2, 2009 #10
    But,

    when we say that a series [tex]\sum b_{n}[/tex] does not converge to [tex]+\infty[/tex] it does not mean that the series converges to a limit b,

    because

    When [tex]\sum b_{n}[/tex] diverges to [tex]+\infty[/tex] ,by definition we have:

    for all ε>0 there exists a natural No N such that for all ,[tex]n\geq N\Longrightarrow[/tex] [itex]\sum_{k=1}^{n}b_{k}\geq\epsilon[/itex]

    and consequently,

    if the series does not diverge to infinity we have:

    there exists an ε>0 and for all natural Nos N there exists an ,[tex]n\geq N[/tex] and [itex]\sum_{k=1}^{n}b_{k}<\epsilon[/itex].

    Does that mean that [tex]\sum b_{n}[/tex] converges to b ??
     
    Last edited: Dec 2, 2009
  12. Dec 2, 2009 #11
    let me make sure i have my facts straight (& practicing my itex-messaging....)

    - a series converges to A if for all [itex]\epsilon > 0[/itex] there is an N with the property that for k>N, [itex]|(\sum_{n=1}^{k}a_{n}) - A| < \epsilon[/itex]

    - & if it doesn't converge it diverges, either to infinity or it alternates forever like (-1)^n because there's no N with the above property (thx to halls for reminding me :redface:)

    - the comparison test says that if [itex]\sum_{n=1}^{\infty}a_n[/itex] converges, and [itex]|b_n| \leq |a_n|[/itex] for all n, then [itex]\sum_{n=1}^{\infty}b_n[/itex] converges. or conversely, if [itex]\sum_{n=1}^{\infty}a_n[/itex] diverges, and [itex]|b_n| \geq |a_n|[/itex] for all n, then [itex]\sum_{n=1}^{\infty}b_n[/itex] diverges.

    - in particular (set [itex]b_n = (1+1/n)a_n[/itex]), if given that [itex]\sum_{n=1}^{\infty}(1+1/n)a_n[/itex] converges, then [itex]\sum_{n=1}^{\infty}a_n[/itex] converges by comparison with [itex]\sum_{n=1}^{\infty}(1+1/n)a_n[/itex], since [itex] |(1+1/n)a_n| \geq |a_n|[/itex] for all n

    - or looking at it the other way, if [itex]\sum_{n=1}^{\infty}a_n[/itex] diverges, then [itex]\sum_{n=1}^{\infty}(1+1/n)a_n[/itex] diverges by comparison with [itex]\sum_{n=1}^{\infty}a_n[/itex] since [itex] |(1+1/n)a_n| \geq |a_n|[/itex] for all n
     
  13. Dec 2, 2009 #12


    Nowhere is given that [itex]\sum_{n=1}^{\infty}(1+1/n)a_n[/itex] converges.

    And i think that the converse of divergence to infinity, is not convergence ,as i tried to show in my post # 10
     
  14. Dec 2, 2009 #13
    a series converges (to a number A) if for all [itex]\epsilon > 0[/itex] there is an N with the property that for k>N, [itex]|(\sum_{n=1}^{k}a_{n}) - A| < \epsilon[/itex] (not what you wrote in post #10) & it diverges if there is no N. The opposite of diverge is converge, which means there is such an N for all epsilon. It doesn't matter whether it diverges to infinity or oscillates forever. & if, as in the original post, you want to use one series to determine convergence or divergence of another there needs to be a way of comparing them, and the usual way, at least if you want to use the comparison test it's whether or not [itex]|b_n| \leq |a_n|[/itex]
     
  15. Dec 2, 2009 #14
    Do you agree that the negation of the above definition implies divergence of the series ?
     
  16. Dec 2, 2009 #15
    that's what I said. It diverges if there's no N.
     
  17. Dec 3, 2009 #16

    But the negation of the definition of convergence is the following :

    For all ,A there exists ε>0 and for all N there exists k>N and
    [itex]|(\sum_{n=1}^{k}a_{n}) - A| \geq\epsilon[/itex].

    Do you agree that this denotes the divergence of the series??
     
  18. Dec 3, 2009 #17
    I think that looks pretty close. I would change the "there exists k>N" to something like "when k>N" or "for all k>N"
     
  19. Dec 3, 2009 #18
    Would you agree then that the denial to the divergence of a series to [tex]+\infty[/tex] is :

    There exists an ε>0 and for all natural Nos N there exists an [tex]n\geq N[/tex] and [itex]\sum_{k=1}^{n}a_{k}) <\epsilon[/itex]

    Provided the definition of a series diverging to [tex]+\infty[/tex] is:

    for all ε>0 there exists a natural No N and for all ,n :[tex]n\geq N\Longrightarrow[/tex] [itex]\sum_{k=1}^{n}a_{k} \geq\epsilon[/itex] ??
     
  20. Dec 3, 2009 #19
    the above "definition" of divergence doesn't necessarily mean it goes to infinity, it means that the partial sums doesn't get closer to that number A.
     
  21. Dec 4, 2009 #20
    How would you define a series that go to [tex]+\infty[/tex]
     
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