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Series Question

  1. Oct 13, 2006 #1
    If we are given a convergent sequence [tex] a_{n} [/tex], then [tex] \frac{1}{a_{n}} [/tex] diverges. But what about the converse:



    If [tex] \frac{1}{a_{n}} [/tex] diverges, then [tex] a_{n} [/tex] converges? [tex] a_{n} \rightarrow 0 [/tex], so its possible that [tex] a_{n} [/tex] could converge. But it could also diverge, right?
     
    Last edited: Oct 13, 2006
  2. jcsd
  3. Oct 13, 2006 #2

    StatusX

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    Take a_n=1.
     
  4. Oct 13, 2006 #3
    ok thanks. so then it could both converge and diverge then.
     
  5. Oct 13, 2006 #4

    quasar987

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    You say "series" in the topic's title but you talk about sequences. And you say something that doesn't make sense to me: in my world, if a_n goes to a[itex]\neq[/itex]0, then 1/a_n goes to 1/a.

    On the other hand if [itex]\sum a_n[/itex] converges, then [itex]\sum \frac{1}{a_n}[/itex] diverges. Is that what you meant?
     
    Last edited: Oct 13, 2006
  6. Oct 13, 2006 #5
    yes sorry for the bad terminology
     
  7. Oct 13, 2006 #6

    quasar987

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    Ok, then in this case, if the sequence [itex]1/a_n[/itex] diverges, then [itex]a_n \rightarrow 0[/itex] and although it is certain that [itex]\sum\frac{1}{a_n}[/itex] diverges, [itex]\sum a_n[/itex] may or may not converges.
     
  8. Oct 13, 2006 #7

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    Is this homework? If not, and you're just messing around to get a better handle on series, then I shouldn't have been so short (although you should probably post questions like this in the general math section). If you want to keep pursuing this line, try comparing the convergence of the sum of {a_n} with that of {(a_n)^p} for values of p other than -1, or even {f(a_n)} for arbitrary functions f. Can you find any conditions on p or f that allow you to conclude the convergence of the latter series from the former, or vice versa. Just a suggestion.
     
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