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Math Help - Question about the root test for series

  1. #1
    Senior Member mollymcf2009's Avatar
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    Question about the root test for series

    Please see the attached worked problem.

    Root Test example.pdf

    My question is about the final limit:

    \lim_{n\rightarrow\infty} [\frac{n}{n} \cdot \frac{(n-1)}{n} \cdot (n-2)!]

    Can someone show me the steps to get to this point? I am confused as to how the limit becomes this. I know that it is because of the factorial, but it this from canceling due to a telescoping series?

    Thanks!
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  2. #2
    o_O
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    So, by simple exponent rules: a_n = \frac{(n!)^n}{n^{2n}} = \frac{(n!)^n}{\left(n^{2}\right)^n} = \left(\frac{n!}{n^2}\right)^n

    Now, to use the root test we consider: \lim_{n \to \infty} \sqrt[n]{\left|a_n\right|} = \lim_{n \to \infty} \sqrt[n]{\left(\frac{n!}{n^2}\right)^n} = \lim_{n \to \infty} \frac{n!}{n^2}

    Recall that: n! \ \ = \ \ n(n-1)! \ \ = \ \ n(n-1)(n-2)! \ \ = \ \ n(n-1)(n-2)\cdots 2 \cdot 1

    So, our limit becomes: \lim_{n \to \infty} \frac{n(n-1)(n-2)!}{n \cdot n} = \lim_{n \to \infty} \left( \frac{n}{n} \cdot \frac{n-1}{n} \cdot (n-2)!\right) (just breaking fraction apart to get rid of the n^2)

    Now, \frac{n}{n} is simply 1 and \lim_{n \to \infty} \frac{n-1}{n} = 1.

    So our limit finally becomes: \lim_{n \to \infty} (n-2)!  = \infty \ \ {\color{red}>} \ \ 1
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