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A limsup inequality (showing that the root test is stronger than the ratio test)

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that if [itex]a_n > 0[/itex] for all n,

    [itex]\liminf{\frac{a_{n+1}}{a_n}} \leq \liminf{a_n^{1/n}} \leq \limsup{a_n^{1/n}} \leq \limsup{\frac{a_{n+1}}{a_n}}[/itex]

    2. Relevant equations

    [itex]\liminf{a_n^{1/n}} \leq \limsup{a_n^{1/n}}[/itex]

    [itex]\liminf{\frac{a_{n+1}}{a_n}} \leq \limsup{\frac{a_{n+1}}{a_n}}[/itex]
    These are obvious, by properties of sup and inf.

    Which implies that, to complete the proof, I must show that

    [itex]\liminf{\frac{a_{n+1}}{a_n}} \leq \liminf{a_n^{1/n}}[/itex]
    [itex]\limsup{a_n^{1/n}} \leq \limsup{\frac{a_{n+1}}{a_n}}[/itex]

    I'm focusing right now on the limsup inequality.

    3. The attempt at a solution

    I took ln of each term,

    [itex]\ln{\limsup{a_n^{1/n}}} = \limsup{\ln{a_n^{1/n}}} = \limsup{\frac{\ln{a_n}}{n}}[/itex]


    [itex]\ln{\limsup{\frac{a_{n+1}}{a_n}}} = \limsup{\ln{\frac{a_{n+1}}{a_n}}} = \limsup{(\ln{a_{n+1}}-\ln{a_n})}[/itex]

    This implies that I need to show that

    [itex]\limsup{\frac{\ln{a_n}}{n}} \leq \limsup{(\ln{a_{n+1}}-\ln{a_n})}[/itex]

    I've let

    [itex]M_1 := \limsup{\frac{\ln{a_n}}{n}}[/itex]
    [itex]M_2 := \limsup{(\ln{a_{n+1}}-\ln{a_n})}[/itex]

    So now I just have to show that [itex]M_1 \leq M_2[/itex]

    By the definition of limit, for all ε>0 there exists an N1 in the natural numbers such that for all [itex]n > N_1[/itex] we have

    [itex]M_1 - ε < \frac{\ln{a_n}}{n} < M_1 + ε[/itex]

    Similarly, for all ε>0 there exists an N2 such that for all [itex]n > N_1[/itex] we have

    [itex]M_2 - ε < \ln{a_{n+1}}-\ln{a_n} < M_2 + ε[/itex]

    I think I can say that the epsilons are the same, since for each of these statements, I can choose a specific epsilon and I'm guaranteed that an N1, N2 exist to satisfy the inequalities. However, I don't know how to go from here to show [itex]M_1 \leq M_2[/itex]. It might be trivial, I don't know, but I just don't see it. I've been working on this problem for too long, and it's making me cross-eyed. It was a homework problem that I turned in unfinished, and now I'm just trying to finally figure it out for my own sanity.
  2. jcsd
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