1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Convergent -divergent series
  1. Convergence of Series (Replies: 17)

  2. Convergent Series (Replies: 3)

Loading...