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    Greetings!

    I'm trying to work out the radius of convergence R of the function

    \sum_{n=1}^{\infty} \frac{n!z^n}{n^n}

    using the Cauchy root test:

    \frac{1}{R}=\displaystyle\lim_{n  \to\infty}\left(\frac{n!}{n^n} \right)^{1/n}.

    To go about solving this, I set y=\frac{n!^{1/n}}{n} and took the natural logarithm of y:

    \ln y = \frac{1}{n}\ln n! - \ln n.

    I then used Stirling's approximation, n!\approx \sqrt{2\pi n} n^n e^{-n} (my mathematician friend laughed at me for doing this, but I'm a physicist and thus love a good approximation ), and so

    \ln y = \frac{1}{n} \left[\frac{1}{2}\ln 2\pi n + n\ln n -n \right]-\ln n,

    which simplifies to

    \ln y = \frac{\ln 2\pi n}{n}-1

    and so taking the limit gives \displaystyle\lim_{n\to\infty} \ln y = -1. This means the answer to the original limit is given by

    \displaystyle\lim_{n\to\infty}y=  \displaystyle\lim_{n\to\infty} \frac{n!^{1/n}}{n}=\frac{1}{e}

    and so the radius of convergence is e (or at least I think it is!).

    The book solution simply states that "Since the nth root of n! tends to n as n \to \infty [poorly worded!], the limit of this ratio is that of \frac{n}{n} , namely unity. Thus R = 1 and the series converges inside the unit circle." I don't really understand where this result comes from, and why it is different to mine! Was my friend right for laughing at my use of Stirling's approximation?

    Any help and insights will be much appreciated Thanks!
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    It looks like it'd be easier with the ratio test.

    EDIT: I get the same answer as you, using the ratio test (much more easily/rigorously). Wolfram agrees with me (well, agrees that R isn't 1), since using z = 2 the series converges.
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    Thanks for the response! Did you use Stirling when you did the ratio test? I'm wondering if there is a different method (which I'm sure there is) of dealing with the n!...
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    (Original post by arkadiy)
    Thanks for the response! Did you use Stirling when you did the ratio test? I'm wondering if there is a different method (which I'm sure there is) of dealing with the n!...
    I really would avoid using things like Stirling if possible. The n! cancels nicely in the ratio test
 
 
 
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