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Math Help - Convergence Problem

  1. #1
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    Convergence Problem

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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    ii) follow the hint which is given...

    iii)
    • x=\frac{1}{4} : Are you allowed to use Stirling's approximation ? (for n\to\infty, n!\sim\sqrt{2\pi n}\left(\frac{n}{\mathrm{e}}\right)^n)
    • x=-\frac{1}{4} : Think about the alternating series test.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by szpengchao View Post
    The second one is asking for

    \sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^2}x^n

    First you do rato test

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

    Since simplifying we would have \lim_{n\to\infty}\bigg|\frac{x\cdot{x^{n}}(2n+2)(2  n+1)(2n)!}{(n+1)^2(n!)^2}\cdot\frac{(n!)^2}{(2n)!x  ^b}\bigg|=\lim_{n\to\infty}|4x|<1\Rightarrow{x<\fr  ac{1}{4}\text{or}x>\frac{-1}{4}}

    Now we can see that \sum_{n=0}^{\infty}\frac{(2n)!(-1)^n}{(n!)^24^n}

    Converges since \lim_{n\to\infty}\bigg|\frac{(2n)!(-1)^n}{(n!)^24^n}\bigg|=\lim_{n\to\infty}\frac{(2n)  !}{(n!)^24^n}=0 and a_{n+1}<a_n

    Now for the other test use stirlings approximation


    I showed you the first part in case you needed the interval of convergence
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    What link is there between " a_n\to_0 for n\to\infty" and " \sum a_n diverges" ?
    Yeah I saw that...I changed it before you posted....I meant convergent by alternating series test
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  5. #5
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    check

    please check if my proof for the first part is right
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by szpengchao View Post
    please check if my proof for the first part is right
    where did that come from?
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  7. #7
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    first step

    the first step is wallis product
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by szpengchao View Post
    \lim_{n\to\infty}\frac{(2n)!}{(n!)^24^n}

    Letting L equal that limit we get

    Take the natural logs of both sides

    and note taht \ln((2n)!)=\sum_{m=1}^{2n}\ln(m)
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  9. #9
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by szpengchao View Post
    please check if my proof for the first part is right
    If you're allowed to use the Wallis product, (Is it in the exercise ?) it's OK.
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  10. #10
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    good good

    just want to check if there are some wrong steps...

    ok...cheers!

    so for the last part, you did ratio test and try to find the radius of convergence, right?

    then the result comes straight from it?
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by szpengchao View Post
    just want to check if there are some wrong steps...

    ok...cheers!

    so for the last part, you did ratio test and try to find the radius of convergence, right?

    then the result comes straight from it?
    Yes I found the radius of convergence and showed that \sum{a_n}

    is convergent \forall{x}\in\bigg(\frac{-1}{4},\frac{1}{4}\bigg)

    Then we msut check the endpoints...through the alternating series test we can see that x=\frac{-1}{4} makes the series converge

    and I leave it up to you to show that the other value makes the series ____
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  12. #12
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    no idea

    i have no idea to show when x=1/4

    please give some hint
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  13. #13
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by szpengchao
    i have no idea to show when x=1/4
    Quote Originally Posted by flyingsquirrel View Post
    • x=\frac{1}{4} : Are you allowed to use Stirling's approximation ? (for n\to\infty, n!\sim\sqrt{2\pi n}\left(\frac{n}{\mathrm{e}}\right)^n)
    ...
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  14. #14
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    yeah

    yeah...i got as n goes to infinity, that is equivalent to 1/sqrt( pi * n)

    so, what's next?
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  15. #15
    Super Member flyingsquirrel's Avatar
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    As for n\to\infty, \left|\binom{2n}{n}\frac{1}{4^n}\right| \sim \frac{1}{\sqrt{\pi n}} and \sum_{n\geq 1}\frac{1}{\sqrt{\pi n}} diverges, (p-series with p=\frac{1}{2}\leq 1) the series \sum_{n\geq 1}\binom{2n}{n}\frac{1}{4^n} diverges too. It's a theorem but I don't know how it is called in English.
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