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Math Help - Complex Sequence Limits

  1. #1
    Junior Member
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    Jan 2010
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    Complex Sequence Limits

    I am asked to prove that if z\in\mathbb{C} then \displaystyle lim_{n\rightarrow\infty} \frac{z^n}{n^n}=0 and \displaystyle lim_{n\rightarrow\infty} \frac{z^n}{n!}=0

    So I assume I have to treat these as sequence because \frac{z^n}{n^n} cannot be a function of z since the n is there. I know the definition of the limit but I just get stuck. In z were real I think I would take the log and proceed from there, but that seems considerably harder with the complex logarithm.

    Anyway I have:

    Let \epsilon >0. Now I have to find a K such that when n>K, |\frac{z^n}{n^n}-0|<\epsilon which I cannot do.

    Can anyone help me... thanks.
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  2. #2
    Junior Member
    Joined
    Jan 2009
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    Well if you want you can use the fact that
    z= r exp(it) where r and t are real and then look on the real sequence of |z|^n/n^n=r^n/n^n, where presumably you have proven it already in your real analysis course.
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