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Analysis - a liminf inequality

  1. Jun 23, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    I'm asked to show that

    [tex]\liminf |a_{n+1}/a_n|\leq\liminf (|a_n|)^{1/n}[/tex]

    3. The attempt at a solution

    I tried a proof by contradiction in both "directions", but it did not work. For instance, one "direction" would be to let [itex]M=\liminf |a_{n+1}/a_n|[/itex] and [itex]L=\liminf (|a_n|)^{1/n}[/tex] and suppose M>L. Then for all e'>0, there is an N' such that n>N' ==>[itex]|a_{n+1}/a_n|>M-\epsilon'[/itex]. Also, let [itex]\{a_{n_k}\}[/itex] be the subsequence such that [itex](a_{n_k})^{1/n_k}\rightarrow L[/itex]. Then, given e>0, there is an N such that k>N ==> [itex]L-\epsilon<(a_{n_k+1})^{1/n_k+1}<L+\epsilon[/itex] and [itex]|a_{n_k+1}/a_{n_k}|>M-\epsilon'[/itex]. But

    [tex]L-\epsilon<(a_{n_k+1})^{1/n_k+1}<L+\epsilon \Leftrightarrow [/tex]

    [tex](L-\epsilon)^{1/n_k+1}<a_{n_k+1}<(L+\epsilon)^{1/n_k+1}\Leftrightarrow[/tex]

    [tex]\frac{(L-\epsilon)^{1/n_k+1}}{a_{n_k}}<\frac{a_{n_k+1}}{a_{n_k}}<\frac{(L+\epsilon)^{1/n_k+1}}{a_{n_k}}\Leftrightarrow[/tex]
    [tex]\frac{(L-\epsilon)^{1/n_k+1}}{(L+\epsilon)^{1/n_k}}<\frac{a_{n_k+1}}{a_{n_k}}<\frac{(L+\epsilon)^{1/n_k+1}}{(L-\epsilon)^{1/n_k}}[/tex]

    [tex]\Rightarrow \frac{a_{n_k+1}}{a_{n_k}}<\left(\frac{L+\epsilon}{L-\epsilon}\right)^{1/n_k}}(L+\epsilon)}[/tex]

    And then I aked wheter we could find an e such that this inequality would contradict [itex]|a_{n_k+1}/a_{n_k}|>M-\epsilon'[/itex], but obviously not, since as n_k gets bigger and bigger, the inequality becomes less restrictive rather than the opposite...


    I also tried the same thing with starting with the fact that [itex]\{a_{n_k}\}[/itex] was the subsequence that converged to M. I arrived at the same kind of uninteresting inequality.

    Using the same ideas, I also tried to contradict the fact that if L is the liminf, then for all N, there is an n>N with (a_n)^(1/n)>L+e (and analogously for M).
     
    Last edited: Jun 23, 2007
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  3. Jun 27, 2007 #2

    quasar987

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    bump!

    This should not be so difficult; it is problem 51 in the book (odd numbered) and there isn't even a hint at the back of the book.

    Also, it may be helpful to know (problem 47) that the ratio and root tests can be generalized as follow: "If limsup|a_{n+1}/a_n|<1, the series converges, and if liminf|a_{n+1}/a_n|>1, the series diverges." (and analogously for (a_n)^1/n).
     
  4. Jul 2, 2007 #3

    quasar987

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    I emailed a professor and he emailed me back a nice pdf with the solution. It involves letting b_n=ln|a_n|, so that the inequality to show is equivalent to showing that liminf(b_n - b_{n-1}) < liminf(b_n/n) and then letting c_n = b_n - b_{n-1}, so it's equivalent to showing liminf(c_n)<liminf((c_1+...+c_n)/n). then let L=liminf(c_n), M=liminf((c_1+...+c_n)/n) and plug the fact that for any e>0, there's N s.t. n>N ==> c_n > L - e into (c_1+...+c_n)/n. It will result by taking the limits that [itex]M\geq L - \epsilon[/itex]. Since this is true for any e, we get the desired [itex]M\geq L[/itex].

    There is a similar inequality for limsup: [tex]\limsup |a_{n+1}/a_n| \geq \limsup (|a_n|)^{1/n}[/tex].

    This is nice because it implies that if the generalized root test is inconclusive, then so will the generalized ratio test so no need to waste time trying it! :smile:
     
    Last edited: Jul 2, 2007
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