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Math Help - series convergence by root test

  1. #1
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    series convergence by root test

    <br />
\left(\frac{k}{k + 1}\right)^{k^2}<br />

    The book answer is confusing to me - I would think you could just take the limit of

    <br />
\frac{k}{k + 1}<br />

    but I am missing something apparently
    Last edited by mr fantastic; April 7th 2009 at 04:55 PM. Reason: Improved the latex
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  2. #2
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    Quote Originally Posted by TYTY View Post
    <br />
\left(\frac{k}{k + 1}\right)^{k^2}<br />

    The book answer is confusing to me - I would think you could just take the limit of

    <br />
\frac{k}{k + 1}<br />

    but I am missing something apparently
    If you apply the Cauchy root test then you need to find the following limit:

    \lim_{k \rightarrow + \infty} \left( \frac{k}{k+1} \right)^k = \lim_{k \rightarrow + \infty} \left( 1 - \frac{1}{k+1} \right)^k.

    It is not too hard to show that the value of this limit is equal to e^{-1} = \frac{1}{e}.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    If you apply the Cauchy root test then you need to find the following limit:

    \lim_{k \rightarrow + \infty} \left( \frac{k}{k+1} \right)^k = \lim_{k \rightarrow + \infty} \left( 1 - \frac{1}{k+1} \right)^k.

    It is not too hard to show that the value of this limit is equal to e^{-1} = \frac{1}{e}.
    re: bolded - maybe for you lol.

    thank you mister fantastic

    I will wrestle with it and also try to figure out why you need

    \lim_{k \rightarrow + \infty} \left( \frac{k}{k+1} \right)^k = \lim_{k \rightarrow + \infty} \left( 1 - \frac{1}{k+1} \right)^k.

    This problem definitely touches on some of my weaker areas. Also thanks for the latex fix, now I learned something new there too
    Last edited by mr fantastic; April 7th 2009 at 05:27 PM. Reason: Fixed the bold tag
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  4. #4
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    Quote Originally Posted by TYTY View Post
    re: bolded - maybe for you lol.

    thank you mister fantastic

    I will wrestle with it and also why you need

    \lim_{k \rightarrow + \infty} \left( \frac{k}{k+1} \right)^k = \lim_{k \rightarrow + \infty} \left( 1 - \frac{1}{k+1} \right)^k.

    This problem definitely touches on some of my weaker areas. Also thanks for the latex fix, now I learned something new there too
    1. \left[ \left( \frac{k}{k+1}\right)^{k^2}\right]^{1/k} = \left( \frac{k}{k+1}\right)^{\frac{k^2}{k}} = \left( \frac{k}{k+1}\right)^{k}.


    2. \frac{k}{k+1} = \frac{(k + 1) - 1}{k+1} = \frac{k+1}{k+1} - \frac{1}{k+1} = 1 - \frac{1}{k+1}.


    3. Use the well known limit from calculus: \lim_{n \rightarrow + \infty} \left(1 + \frac{a}{n} \right)^n = e^a.
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  5. #5
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    You're the man!

    many many thanks
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