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Purely epsilon-N proof of the Leibniz criterion for alternate series

  1. Jun 3, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    I found a purely epsilon-N proof of the Leibniz criterion for alternate series and it is quite inelegant compared to the classical proof so no wonder I never saw it in any textbook. But at the same time I must wonder if I made a mistake somewhere.

    The statement of the Leibniz criterion for alternate series is that if we have an alternating series [itex]\sum a_n[/itex] with [itex]a_n \rightarrow 0[/itex] and that is decreasing in absolute values ([itex]|a_{n+1}\leq |a_n|[/itex]), then the sum converges.


    3. The attempt at a solution I begin by arguing that if for some N, |a_N|=0, then a_n=0 for all subsequent n also. (Because [itex]0\leq |a_{N+k}|\leq |a_N|=0 \Leftrightarrow |a_{N+k}| = 0[/itex]). So if this is the case, then the series converges to [itex]\sum_{n=1}^{N}a_n[/itex]. If a_n is never 0, then we can write

    [tex]0< \frac{|a_{n+1}|}{|a_n|}\leq 1[/tex]

    and therefor, by the "sandwich lemma",

    [tex]0\leq \lim_{n\rightarrow \infty}\frac{|a_{n+1}|}{|a_n|}\leq 1[/tex]

    In the case where the limit is lesser than 1, then the series converges absolutely by the ratio test. In the case where the limit equals 1, the ratio test does not allow us to conclude so we must work a little.

    The fact that the limit is 1 means that for all [itex]\epsilon>0[/itex], there is a N s.t.

    [tex]n\geq N \ \Rightarrow |\frac{|a_{n+1}|}{|a_n|}-1|<\epsilon \Leftrightarrow 1-\epsilon < \frac{|a_{n+1}|}{|a_n|}<1+\epsilon \Leftrightarrow 1-\epsilon < -\frac{a_{n+1}}{a_n}<1+\epsilon[/tex]

    (Because {a_n} is alternating)

    [tex]\Leftrightarrow -1-\epsilon<\frac{a_{n+1}}{a_n}<\epsilon-1 \Leftrightarrow -a_n(\epsilon+1)<a_{n+1}<a_n(\epsilon-1)[/tex].

    In particular, if we chose [itex]\epsilon \in (1,2)[/itex], we get [itex](\epsilon-1)\in (0,1)[/itex] and the following "identities":

    [tex]-a_N(\epsilon+1)<a_{N+1}<a_N(\epsilon-1) \ \ \ \ (1)[/tex]

    [tex]-a_{N+1}(\epsilon+1)<a_{N+2}<a_{N+1}(\epsilon-1)[/tex]

    But from equation (1), [itex]-a_{N+1}>-a_{N}(\epsilon-1)[/itex] and [itex]a_{N+1}<a_N(\epsilon-1)[/itex], so

    [tex]-a_{N}(\epsilon-1)(\epsilon+1)<a_{N+2}<a_{N}(\epsilon-1)^2 \ \ \ \ (2)[/tex]

    [tex]-a_{N+2}(\epsilon+1)<a_{N+3}<a_{N+2}(\epsilon-1)[/tex]

    But from equation (2), [itex]-a_{N+2}>-a_{N}(\epsilon-1)^2[/itex] and [itex]a_{N+2}<a_N(\epsilon-1)^2[/itex], so

    [tex]-a_{N}(\epsilon-1)^2(\epsilon+1)<a_{N+3}<a_{N}(\epsilon-1)^3[/tex]

    Etc., by induction, we get that

    [tex]-a_N(\epsilon-1)^{k-1}(\epsilon+1)<a_{N+k}<a_{N}(\epsilon-1)^k[/tex]

    Thus, again by the sandwich lemma, the (truncated) alternate series is squeezed btw two geometric series:

    [tex]-a_N(\epsilon+1)\sum_{n=0}^{\infty}(\epsilon-1)^n\leq \sum_{n=N+1}^{\infty}a_n\leq a_N\sum_{n=1}^{\infty}(\epsilon-1)^k[/tex]


    Did you notice that I never use the hypothesis [itex]a_n\rightarrow 0[/itex]? Not good. Or did I unknowingly?
     
    Last edited: Jun 3, 2007
  2. jcsd
  3. Jun 4, 2007 #2
    The RHS of your inequality doesn't even converge, unless you've made a typo.
     
  4. Jun 4, 2007 #3

    matt grime

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    This does not follow. One of a_n is positive, and one is negative, you don't know which, and can't assume it is a_n that is -ve
     
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