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Divergence of an infinite series (using the def of limit)

  1. Oct 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Given that [tex]a_{n} > 0[/tex] and [tex]lim(na_{n}) = l[/tex] with [tex]l\neq0[/tex],
    prove that [tex]\sum a_{n}[/tex] diverges.


    2. Relevant equations



    3. The attempt at a solution

    lim(na_n)=l (with =/= 0), so I can safely say that:

    [tex]\left|na_{n}-l\right| < \epsilon[/tex] by the definition of limit.

    Then isn't it also true that [tex]\left|a_{n}-l\right| < \epsilon[/tex] because [tex]\left|a_{n}-l\right| \leq \left|na_{n}-l\right|[/tex] and is smaller than the same epsilon?

    From there it would imply that a_n converges to l which is never 0, so the sum of a_n would not converge either.
     
    Last edited: Oct 24, 2009
  2. jcsd
  3. Oct 24, 2009 #2

    Office_Shredder

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    In fact, you expect |an-l| to be approximately l for large values of n. Why? Because nan converges to a value, so an must go to zero since n goes to infinity
     
  4. Oct 24, 2009 #3
    I made a mistake on the original post. The goal is to prove that [tex]\sum a_{n}[/tex] diverges. Sorry about that... Does my explanation make more sense now?
     
  5. Oct 24, 2009 #4

    Office_Shredder

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    This part is true

    This part isn't. For example, if an=1/n, then nan converges to 1, but an converges to 0!
     
  6. Oct 24, 2009 #5
    ok, thank you. I think I proved it by letting epsilon = 2l (which is >0 because a_n >0)

    Then [tex]-2l < na_{n}-l < 2l[/tex] implies [tex]-l < na_{n} < 3l[/tex], which implies [tex]-l/n < a_{n}[/tex]

    Since ∑ -l/n = (-1)*∑ 1/n which doesn't converge, by comparison test a_n doesn't converge.
     
  7. Oct 24, 2009 #6

    Office_Shredder

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    You have that -l/n<an<3l/n

    This doesn't tell you anything! an could be, for example, 1/n2 and satisfy this. The comparison test only works when all the terms in both series are positive

    When you have

    |nan-l|<e, what can you say about |an-l/n|?
     
  8. Oct 24, 2009 #7
    |an-l/n| approaches 0 doesn't it? I'm not sure if this was what you were getting at... I understand that an is some form of 1/n, but I want to prove that by using the e def of limit and without providing a specific an that works.

    How about if e = l/2? Then I would have l/2 < an < l/2 + l. Couldn't I use the comparison test on l/2 < an? Thanks again.
     
  9. Oct 24, 2009 #8

    Office_Shredder

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    Stop trying to bound it below by a single number. You know an goes to zero, so if you can bound it below by l/2, you have a contradiction. I don't see how you got that last inequality

    |an-l/n| goes to zero, but how?

    Look:
    [tex] |na_n-l|< \epsilon \rightarrow |a_n-\frac{l}{n}|< \frac{ \epsilon}{n}[/tex]

    So what can we say about an? For large enough n, this:

    [tex] \frac{l}{n} - \frac{ \epsilon}{n} < a_n < \frac{l}{n} + \frac{ \epsilon}{n}[/tex]

    So what value of epsilon do you want, and what series are you comparing to?
     
  10. Oct 24, 2009 #9
    If [tex]\epsilon = \frac{l}{2}[/tex], we have [tex]\left| na_{n} - l \right| < \epsilon = \frac{l}{2}[/tex]

    which [tex]= -\frac{l}{2} < na_{n} - l < \frac{l}{2}[/tex]

    [tex]= -\frac{l}{2n} < a_{n} - l < \frac{l}{2n}[/tex]

    [tex]= -\frac{l}{2n} + l < a_{n} < \frac{l}{2n} + l[/tex]

    [tex]= \frac{l}{2n} < a_{n} < \frac{l}{2n} + l[/tex]
     
  11. Oct 24, 2009 #10


    Can [tex]\epsilon = \frac{l+\epsilon}{n}[/tex]?...
     
  12. Oct 24, 2009 #11

    Office_Shredder

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    Probably not...

    You had the right idea with comparison test. You just need to figure out what value of epsilon gives you something useful
     
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