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Proving part of the ratio test

  1. Jun 2, 2014 #1
    This is not a homework problem. I'm doing it for fun. But it is the kind that might appear on homework.

    1. The problem statement, all variables and given/known data

    I'm trying to prove that if lim n→∞ |an+1/an| = L < 1, then [itex]\Sigma[/itex] an converges absolutely and therefore converges.


    2. Relevant equations



    3. The attempt at a solution

    Here's my thinking. I feel like I'm on the right track, but I may need some help formalizing my expression of what's happening "in the limit" as n→∞.

    To show that ∑an converges absolutely, I need to show that ∑|an| converges. My strategy is to show that, because of the condition above, as n→∞ this "tends toward" a geometric series with common ratio < 1 and therefore converges.

    I have

    ∑an = |a1| + |a2| + |a3| + ... + |an|

    = |a1| + |a1||a2/a1| + |a1||a2/a1||a3/a2| + ... + |a1||a2/a1||a3/a2||a4/a3|...|an/an-1|

    Supposing lim n→∞ |an+1/an| = L < 1, this means that there is some integer N such that for n > N, all the ratios |an+1/an| are equal in the sense that the difference between any two of them can be made arbitrarily small by choosing N appropriately.

    Therefore I conclude that for n > N, the series above can be written

    ∑|an| = |a1|(1 + L + L2 + L3 + L4 + ... + Ln)

    i.e. it is a geometric series (or eventually becomes one beyond N) with common ratio < 1 and therefore converges.

    And finally, since ∑|an| converges, ∑an converges absolutely and therefore it also converges.

    Is my reasoning solid here? I am particularly concerned about the conclusion that all the ratios in the series (the ones that multiply |a1| in each term) eventually equal L. I think my intuition here is correct, but I am not well enough attuned to the subtleties of analysis to be confident that my chain of reasoning is unimpeachable.

    NOTE: I know there must be better, shorter, more elegant, less cumbersome ways of proving this. I'm not interested in those until I have developed my own proof to the utmost. That way I will learn the most from the process.

    Thanks!
     
  2. jcsd
  3. Jun 2, 2014 #2
    This is what troubles me most. A bit hand-wavey, I fear. I am wondering if I shouldn't try to show this explicitly using epsilonic reasoning, etc.
     
  4. Jun 2, 2014 #3
    Rather than showing your series "tends toward" a geometric series, maybe try to show that it is eventually smaller than a convergent geometric series.
     
  5. Jun 2, 2014 #4
    In undergraduate analysis, most proofs require an epsilon. This is one of them.
     
  6. Jun 2, 2014 #5

    BiGyElLoWhAt

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    hmmm... is that a proper representation of the ratio test?

    What about something like this:

    ##∑(5-\frac{n}{1000})##

    Or is this a situation where the ratio test doesn't apply? I'm just asking, because this fits within your conditions, ##\frac{a_{n+1}}{a_{n}}<1## but as n approaches infinity, ##a_{n}→5##

    I think you need:
    if
    ##\frac{a_{n+1}}{a_{n}}<1## and ##\displaystyle\lim_{n→\infty}a_{n}=0## then S is convergant.

    I know this doesn't really help you answer your question, but I feel it's a necessary condition.
     
  7. Jun 2, 2014 #6

    BiGyElLoWhAt

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    oh wait, I lied. ##a_{n}→-\infty##
    either way, it doesn't converge...
     
  8. Jun 2, 2014 #7

    jbunniii

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    No, that's not enough to ensure convergence. If ##a_n = 1/n## then ##a_n \rightarrow 0##
    $$\frac{1/(n+1)}{1/n} = \frac{n}{n+1} < 1$$
    but ##\sum a_n## diverges.

    I suggest the following approach. We are given that
    $$\lim \left|\frac{a_{n+1}}{a_n}\right| = L < 1$$
    Therefore if we pick some number ##M## satisfying ##L < M < 1##, then there is some ##N## such that
    $$\left|\frac{a_{n+1}}{a_n}\right| < M$$
    for all ##n \geq N##. Now use the fact that if ##n > N##,
    $$|a_n| = \left|\frac{a_{n}}{a_{n-1}}\right|\cdots\left|\frac{a_{N+1}}{a_N}\right||a_N|$$
     
  9. Jun 2, 2014 #8

    pasmith

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    No, [itex]5 - \frac{n}{1000} \to -\infty[/itex]. Also, the condition is that [tex]
    \lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|} < 1[/tex], and for [itex]a_n = 5 - n/1000[/itex] we get [tex]
    \frac{a_{n+1}}{a_n} = \frac{n + 1 - 5000}{n - 5000} \to 1[/tex] so this is not a counter-example.

    In fact, if [itex]|a_{n+1}|/|a_n|[/itex] tends to anything other than 1 then it must be the case that [itex]|a_n| \to 0[/itex]: If [itex]|a_n| \to a > 0[/itex] then also [itex]|a_{n+1}| \to a > 0[/itex] and thus [itex]|a_{n+1}|/|a_{n}| \to a/a = 1[/itex] by the theorem that if the limits of two sequences exist, then the limit of the product exists and is the product of the limits. But if [itex]|a_n| \to 0[/itex] then the sequence [itex]1/|a_{n}|[/itex] diverges to [itex]+\infty[/itex] and that theorem doesn't apply: the limit of [itex]|a_{n+1}|/|a_{n}|[/itex] is in this case of the indeterminate form [itex]0/0[/itex].
     
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