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Math Help - Limit infina of ratio test and root test

  1. #1
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    Limit infina of ratio test and root test

    For a strictly positive sequence, prove that \liminf (a_j^{\frac{1}{j}}) \geq \liminf (\frac{a_{j+1}}{a_j}).

    I get that it's trying to say that the root test converges faster than the ratio test (hence the greater limit inf), but beyond that, I am drawing a blank. I don't even know where to begin.
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  2. #2
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    Re: Limit infina of ratio test and root test

    Hint: For x>0, \lim_{n\to \infty} \sqrt[n]{x}=1. In particular, \liminf_{n\to \infty} \sqrt[n]{x}=\lim_{n\to \infty} \sqrt[n]{x}=1.

    Now think about why \liminf \left(\frac{a_{n+1}}{a_n}\right)\leq 1. (It may be useful to also think about why \limsup \left(\frac{a_{n+1}}{a_n}\right) \geq 1). In particular, what happens if the sequence is eventually constant? What if it never converges?

    As an aside, this:
    I get that it's trying to say that the root test converges faster than the ratio test (hence the greater limit inf)
    is not the way to think about it. I'm not even sure if makes sense. Anyways, the following is true for a_n >0:

    \liminf \left(\frac{a_{n+1}}{a_n}\right)\quad\leq\quad \liminf \sqrt[n]{a_n}\quad\leq \quad \limsup \sqrt[n]{a_n} \quad \leq\quad \limsup\left(\frac{a_{n+1}}{a_n}\right)
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  3. #3
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    Re: Limit infina of ratio test and root test

    I still do not understand. At all.
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  4. #4
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    Re: Limit infina of ratio test and root test

    Do you understand the definition of liminf? Can you see that if a sequence a_n converges, liminf a_n = lim a_n = limsup a_n?
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  5. #5
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    Re: Limit infina of ratio test and root test

    Infinum means greatest lower bound. I.e., if A = (0,1), inf A = 0, because 0 <= all elements in A, and 0 is the greatest possible number that satisfies that inequality.

    Lim inf x_n = lim (n -> infinity) {a_n, a_n+1, a_n+2, ...}. Basically, lim inf is the infinum of the sequence infinitely far from the starting element.

    All of that I get. What I don't get is how to put it together in a meaningful way on this proof.
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  6. #6
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    Re: Limit infina of ratio test and root test

    If the sequence  a_n is not eventually constant, there are infinitely many points where a_{n+1} < a_n. Then \limsup \frac{a_{n+1}}{a_n}\leq 1 = \liminf \sqrt[n]a_n = \lim \sqrt[n]a_n. The case where a_n is eventually constant is obvious.
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