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Math Help - convergence problem

  1. #1
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    convergence problem

    \sum \frac{2^{n-1}n!}{n^{2n+1}}
    so, i should use D'Alambert rule. Then i got
    \frac{a_{n+1} }{a_{n}}= \frac{\frac{2^n(n+1)!}{(n+1)^{2n+3}}}{\frac{2^{n-1}n!}{n^{2n+1}}}

    and what now?
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  2. #2
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    Re: convergence problem

    Acknowledging that

    \frac{(n+1)!}{n!} = n+1

    and

    \frac{2^{n}}{2^{n-1}} = 2

    Your ratio can be simplified to

    \lim_{n \to \infty} \frac{a_{n+1}}{a_n}

    = \lim_{n \to \infty} \frac{2(n+1)\cdot n^{2n+1}}{(n+1)^{2n+3}}

    = \lim_{n \to \infty} \frac{2(n+1)\cdot n^{2n+1}}{(n+1)^{2n+1}\cdot (n+1)^2}

    = \lim_{n \to \infty} \left (\frac{n}{n+1}\right )^{2n+1} \cdot \frac{2}{n+1}

    = \lim_{n \to \infty} \frac{2}{\left (1+\frac{1}{n}\right )^{2n+1} \cdot (n+1)}

    = \lim_{n \to \infty} \frac{1}{\left (1+\frac{1}{n}\right )^{2n}} \cdot \lim_{n \to \infty} \frac{1}{1+\frac{1}{n}} \cdot \lim_{n \to \infty} \frac{2}{n+1}

    = \frac{1}{e^2} \cdot 1 \cdot 0

    = 0

    So the series converges.
    Last edited by SworD; September 13th 2012 at 07:29 PM.
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  3. #3
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    Re: convergence problem

    Quote Originally Posted by DonnieDarko View Post
    \sum \frac{2^{n-1}n!}{n^{2n+1}}
    so, i should use D'Alambert rule. Then i got
    \frac{a_{n+1} }{a_{n}}= \frac{\frac{2^n(n+1)!}{(n+1)^{2n+3}}}{\frac{2^{n-1}n!}{n^{2n+1}}}

    and what now?
    Simplify.

    = \left( \frac{2^n}{2^{n-1}} \right) \left( \frac{(n+1)!}{n!} \right) \left( \frac{n^{2n+1}}{(n+1)^{2n+3}} \right) = 2(n+1) \left( \left( \frac{n}{(n+1)^3} \right) \left( \frac{n}{n+1} \right)^{2n} \right)

    = \frac{2n}{(n+1)^2} \left( \frac{n}{n+1} \right)^{2n} = \frac{2n}{(n+1)^2} \left( \frac{n+1-1}{n+1} \right)^{2n} = \frac{2n}{(n+1)^2} \left( 1 - \frac{1}{n+1} \right) ^{2n}

    = \frac{2n}{(n+1)^2} \left( \left( 1 - \frac{1}{n+1} \right) ^{n+1} \right) ^ {\frac{2n}{n+1}}.

    But  \left( 1 - \frac{1}{n+1} \right)^{n+1} \rightarrow e^{-1}, \frac{2n}{n+1} \rightarrow 2, and \frac{2n}{(n+1)^2} \rightarrow 0, so

     \ \frac{a_{n+1}}{a_n} = \frac{2n}{(n+1)^2} \left( \left( 1 - \frac{1}{n+1} \right) ^{n+1} \right) ^ {\frac{2n}{n+1}} \rightarrow (0)(e^{-1})^2 = 0 as  \ n \rightarrow \infty.
    Last edited by johnsomeone; September 13th 2012 at 07:55 PM.
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  4. #4
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    Re: convergence problem

    johnsomeone, your post contains an error. The term with the 2n+3 exponent should be in the denominator.
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  5. #5
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    Re: convergence problem

    Thanks. It's hard to "see" algebra in LaTex!
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  6. #6
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    Re: convergence problem

    Thanks you guys a lot
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