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Divergence of a series

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose [tex](a_n)[/tex] is a sequence of non-negative real numbers such that the series [tex]{\sum_{n=1}}^\infty a_n [/tex] diverges. Prove that the series [tex]{\sum_{n=1}}^\infty \frac{a_n}{1+a_n} [/tex] must also diverge.


    2. Relevant equations



    3. The attempt at a solution

    I was thinking about looking at [tex]l=limsup(a_n) [/tex] and perhaps the requirements on it in the root test in order to see if that could tell me something about the [tex]l=limsup \left( \frac{a_n}{1+a_n} \right) [/tex], but I haven't had much luck.

    Any suggestions?
    Thanks!
     
  2. jcsd
  3. Oct 22, 2009 #2
    Might it help to write

    [tex]\frac{a_n}{a_n + 1} = \frac{a_n + 1 - 1}{a_n + 1} = 1 - \frac{1}{a_n + 1}[/tex]
    ?
     
  4. Oct 22, 2009 #3
    It's kind of a backdoor route, but I'd prove the contrapositive and use a simple comparison test.
     
  5. Oct 22, 2009 #4
    Since [tex]a_n [/tex] is a non negative sequence can we assume that it diverges to [tex]+ \infty[/tex] so then [tex]1 - \frac{1}{a_n + 1} [/tex] diverges to [tex]- \infty[/tex]?
     
  6. Oct 22, 2009 #5
    You're not correct with the limit of the fraction part, but that part isn't really important. What is [itex]\sum_{n = 1}^\infty 1[/itex]?
     
  7. Oct 22, 2009 #6
    Oo oops! It's the lim (1 - 0) = 1. But then wouldn't the sequence then converge to 1 rather than diverging?
     
  8. Oct 22, 2009 #7
    It doesn't converge to 1.
    [tex]\sum_{n = 1}^\infty 1 = ~?[/tex]

    What do you get when you add an infinite number of 1's?
     
  9. Oct 22, 2009 #8
    Wow, tonight is not my night. haha

    [tex]
    \sum_{n = 1}^\infty 1 = + \infty
    [/tex]

    Great! Thanks!
     
  10. Oct 22, 2009 #9

    Office_Shredder

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    Where did this assumption that an goes to infinity come from?
     
  11. Oct 22, 2009 #10
    [tex]\sum_{n=1}^\infty \left(1 - \frac{1}{a_n + 1}\right) = \sum_{n=1}^\infty 1 - \sum_{n=1}^\infty \frac{1}{a_n + 1}[/tex] should diverge whether or not an goes to infinity, no? If it doesn't go to infinity, shouldn't that make the series diverge "more"?
     
  12. Oct 23, 2009 #11

    Office_Shredder

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    Except that if an is small, both series are wildly divergent and you can't make that split in the first place
     
  13. Oct 23, 2009 #12
    hmmm. Is there another way I should approach this problem?
     
  14. Oct 23, 2009 #13
    You can assume that [tex] \frac{a_n}{a_n + 1} [/tex] goes to 0 for n going to infinity, because if it didn't then the series would diverge anyway. [tex] \frac{a_n}{1 + a_n} = 1 - \frac{1}{1 + a_n} [/tex]. If [tex] \lim_{n \rightarrow \infty} 1 - \frac{1}{1 + a_n} = 0 [/tex] then we see that [tex] a_n \rightarrow 0 [/tex] as well.

    For large enough n, [tex] (a_n)^2 < a_n \Rightarrow (a_n)^2 + a_n < 2a_n [/tex].

    Play around with that inequality to get a lower bound for [tex] \frac{a_n}{1 + a_n} [/tex] so that the series for that lower bound diverges.
     
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