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Math Help - Radius of Convergence

  1. #1
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    Radius of Convergence

    Find the radius of convergence of the following series

    \sum_1^\infty \frac{n!}{n^n} t^n

    Using the ratio test I get the following:

    |t| \lim (n\to\infty) |\frac{n+1}{n \cdot n!}

    As far as I can see this converges for all real numbers, but the answer says
    -e<t<e

    Where am I going wrong? Thanks.
    Last edited by Carbon; May 18th 2011 at 03:47 PM. Reason: typo in index
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  2. #2
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    Quote Originally Posted by Carbon View Post
    Find the radius of convergence of the following series
    \sum_0^\infty \frac{n!}{n^n} t^n
    Here are a couple of comments.
    This sum cannot have an index n=0!

    AND \left( {\sqrt[n]{{\frac{{n!}}{{n^n }}}}} \right) \to \frac{1}{e}
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  3. #3
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    Hmmm... thanks. I can see how that would work at the end but I don't see how I can coax this series into being raised to the (1/n) power.
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  4. #4
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    Find |t|<\frac{1}{e}~.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Find |t|<\frac{1}{e}~.
    I don't understand. Didn't you just state what t is equal to? -1/e<t<1/e? I don't understand how to get the power of 1/n from the ratio test in this instance.
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  6. #6
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    Quote Originally Posted by Carbon View Post
    I don't understand. Didn't you just state what t is equal to? -1/e<t<1/e? I don't understand how to get the power of 1/n from the ratio test in this instance.
    Do you understand the root test ?
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  7. #7
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    I guess I don't.

    (All I know is my DE professor is trying to refresh us on the ratio test so that we can learn to solve DEs with power series. I guess I will just hope there's an easier one that doesn't involve e if it comes up on the test. If not, I'll be screwed.)
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  8. #8
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    Quote Originally Posted by Carbon View Post
    I guess I will just hope there's an easier one that doesn't involve e if it comes up on the test. If not, I'll be screwed.)
    You need to know this: \lim _{n \to \infty } \left( {1 + \frac{a}{{n + b}}} \right)^{cn}  = e^{ac}~.

    The ratio yields: \frac{{a_{n + 1} }}{{a_n }} = \left( {\frac{n}{{n + 1}}} \right)^n  = \left( {1 + \frac{{ - 1}}{{n + 1}}} \right)^n

    So you see the limit is e^{-1}
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  9. #9
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    Quote Originally Posted by Plato View Post
    The ratio yields: \frac{{a_{n + 1} }}{{a_n }} = \left( {\frac{n}{{n + 1}}} \right)^n
    That is the part I'm not getting.

    \frac {a_{n+1}}{a_n}}=\frac {\frac{(n+1)!t^{n+1}}{n^{n+1}}}{\frac{n!t^n}{n^n}}  }=\frac {\frac{(n+1)!t^nt}{n^nn}}{\frac{n!t^n}{n^n}}}= \frac {(n+1)!t}{n!n}=\frac{(n+1)t}{n}

    How does one get to a situation where they can raise it to the power of n?
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  10. #10
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    Quote Originally Posted by Carbon View Post
    That is the part I'm not getting.

    \frac {a_{n+1}}{a_n}}=\frac {\frac{(n+1)!t^{n+1}}{n^{n+1}}}{\frac{n!t^n}{n^n}}  }=\frac {\frac{(n+1)!t^nt}{n^nn}}{\frac{n!t^n}{n^n}}}= \frac {(n+1)!t}{n!n}=\frac{(n+1)t}{n}

    How does one get to a situation where they can raise it to the power of n?
    Forget the t, it has nothing to do with the ratio.
    You simply do not know how to do algebra!
    So go back to algebra I. What is you problem?
    I think your problem is simple algebra.
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  11. #11
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    Quote Originally Posted by Plato View Post
    Forget the t, it has nothing to do with the ratio.
    You simply do not know how to do algebra!
    So go back to algebra I. What is you problem?
    I think your problem is simple algebra.
    OK, I know I'm not good at algebra but I'm asking for help. I guess I can repost this in Algebra if that is what I need to do to get help with it...
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  12. #12
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    Quote Originally Posted by Carbon View Post
    That is the part I'm not getting.

    \frac {a_{n+1}}{a_n}}=\frac {\frac{(n+1)!t^{n+1}}{n^{n+1}}}{\frac{n!t^n}{n^n}}  }=\frac {\frac{(n+1)!t^nt}{n^nn}}{\frac{n!t^n}{n^n}}}= \frac {(n+1)!t}{n!n}=\frac{(n+1)t}{n}

    How does one get to a situation where they can raise it to the power of n?
    You have made a slight algebra error. You forgot an n+1

    \frac{a_{n+1}}{a_n}=\frac{\frac{(n+1)!}{(n+1)^{n+1  }}}{\frac{n!}{n^n}}=\frac{(n+1)n!}{(n+1)(n+1)^n}\c  dot \frac{n^n}{n!}=\frac{n^n}{(n+1)^n}=\frac{1}{\left(  1+\frac{1}{n} \right)^n}
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  13. #13
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    Thumbs up

    Quote Originally Posted by TheEmptySet View Post
    You have made a slight algebra error. You forgot an n+1

    \frac{a_{n+1}}{a_n}=\frac{\frac{(n+1)!}{(n+1)^{n+1  }}}{\frac{n!}{n^n}}=\frac{(n+1)n!}{(n+1)(n+1)^n}\c  dot \frac{n^n}{n!}=\frac{n^n}{(n+1)^n}=\frac{1}{\left(  1+\frac{1}{n} \right)^n}
    Thank you!
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