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Thread: Infinite series : convergence.

  1. #1
    MHF Contributor arbolis's Avatar
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    Infinite series : convergence.

    $\displaystyle \sum_{n=1}^{+\infty} \frac{2^nn!}{n^n}$. I must show whether it converges or not. Using my intuition I think it converges. I tried to use the root test, the only test I think that works.
    So I'm stuck when I must find the limit when $\displaystyle n$ tends to $\displaystyle +\infty$ of $\displaystyle \frac{2\cdot n!^{\frac{1}{n}}}{n}$. Does the limit of $\displaystyle n!^\frac{1}{n}=\frac{1}{e}$, when $\displaystyle n$ tends to $\displaystyle +\infty$? If yes, how to prove it using maths of calculus II?
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    $\displaystyle \sum_{n=1}^{+\infty} \frac{2^nn!}{n^n}$. I must show whether it converges or not. Using my intuition I think it converges. I tried to use the root test, the only test I think that works.
    So I'm stuck when I must find the limit when $\displaystyle n$ tends to $\displaystyle +\infty$ of $\displaystyle \frac{2\cdot n!^{\frac{1}{n}}}{n}$. Does the limit of $\displaystyle n!^\frac{1}{n}=\frac{1}{e}$, when $\displaystyle n$ tends to $\displaystyle +\infty$? If yes, how to prove it using maths of calculus II?
    Root test.

    And just consider that all this is limit is is

    $\displaystyle \lim_{n\to\infty}(2^n)^{\frac{1}{n}}\cdot\lim_{n\t o\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}= 2\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{\fr ac{1}{n}}$

    The last limit we have gone over.
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    $\displaystyle \left( {\sqrt[n]{{\frac{{2^n n!}}{{n^n }}}}} \right) \to \frac{2}{e}$
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    $\displaystyle \sum_{n=1}^{+\infty} \frac{2^nn!}{n^n}$. I must show whether it converges or not. Using my intuition I think it converges. I tried to use the root test, the only test I think that works.
    So I'm stuck when I must find the limit when $\displaystyle n$ tends to $\displaystyle +\infty$ of $\displaystyle \frac{2\cdot n!^{\frac{1}{n}}}{n}$. Does the limit of $\displaystyle n!^\frac{1}{n}=\frac{1}{e}$, when $\displaystyle n$ tends to $\displaystyle +\infty$? If yes, how to prove it using maths of calculus II?
    you could just say that $\displaystyle \frac{2^nn!}{n^n}\sim\left(\frac{2}{e}\right)^n$

    By stirlings approximation
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    the ratio test also works nicely here. you end up with the limit being $\displaystyle \frac 2e$ as well, which is less than 1, of course. it is easier to see the limit using this way, in my humble opinion
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    MHF Contributor arbolis's Avatar
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    Thanks to all!! This means that $\displaystyle \sum_{n=1}^{+\infty} \frac{3^nn!}{n^n}$ diverges. (I changed the 2 for a 3). As I have to determine it as well, it's already done!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by arbolis View Post
    Thanks to all!! This means that $\displaystyle \sum_{n=1}^{+\infty} \frac{3^nn!}{n^n}$ diverges. (I changed the 2 for a 3). As I have to determine it as well, it's already done!
    yes
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