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Proof of the ratio test

  1. Nov 7, 2012 #1
    I am trying to understand something in the proof of the ratio test for series convergence.
    If [itex]a_{n}[/itex] is a sequence of positive numbers, and that the ratio test shows that [itex] \lim_{n→∞}\frac{a_{n+1}}{a_{n}} = r < 1[/itex], then the series converges.

    Apparently, the proof defines a number R : r<R<1, and then shows that there exists a N>0 such that [itex]\frac{a_{n+1}}{a_{n}} < R [/itex] for all n>N. It need not to be true in the case where n=N, right? Up to this part I get.

    But then it concludes from the above that, there exists a positive N such that
    [itex] a_{N+1}<a_{n}R [/itex] which does not follow due to the statement in bold.

    Could someone please point out where I am wrong so I can continue this theorem without any qualms? Thanks!

  2. jcsd
  3. Nov 7, 2012 #2


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    Education Advisor

    Hmm, think about what it means for the limit to exist. I think that might be the missing piece that shows why it works. Because if you believe it exist, there is an n greater than or equal to N such that (a_(n+1))/(a_n) is less than or equal to R for n greater than N. Then from there you can rewrite the inequality to get what you got.
  4. Nov 7, 2012 #3
    Are you sure the part in bold is correct? Doesn't the limit definition exclusively use greater than? Because that is precisely what I don't fully understand.

  5. Nov 7, 2012 #4
    It doesn't really matter. The statement:

    For all [itex]\varepsilon>0[/itex], there exists an N such that for all [itex]n\geq N[/itex] holds that [itex]|a_n-a|<\varepsilon[/itex].

    is actually equivalent with

    For all [itex]\varepsilon>0[/itex], there exists an N such that for all [itex]n> N[/itex] holds that [itex]|a_n-a|<\varepsilon[/itex].

    So you can use both statements to define limit of a sequence. Of course, once you decided on which of both versions to use, you have to be consistent.
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