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Math Help - Series converge or diverge?

  1. #1
    Rui
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    Series converge or diverge?

    Use what test?

    a) ∑(from n=1 to infinity) (arctan n/ n^1.2)
    b) ∑(from n=1 to infinity) ( (n^n) / (2n)! )

    Not sure how to do these...when i do it using ratio test, it turns out really weird..so yea i need help.

    Thanks.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by Rui View Post
    Use what test?

    a) ∑(from n=1 to infinity) (arctan n/ n^1.2)
    <br />
\forall x\in \mathbb{R},\,\,|\arctan x|<\frac{\pi}{2} : you may try the comparison test

    b) ∑(from n=1 to infinity) ( (n^n) / (2n)! )
    Ratio test : \left|\frac{a<br />
_{n+1}}{a_n}\right|=\frac{(n+1)^{n+1}}{(2(n+1))!}\  cdot \frac{(2n)!}{n^n}=\frac{(n+1)^{n+1}}{n^n(2n+1)(2n+  2)} because (2(n+1))!=(2n+2)!=(2n+2)\cdot(2n+1)\cdot(2n)!.

    Then, you can use n+1<n to say that (n+1)^{n+1}<n^{n+1} and see what it gives.
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by Rui View Post
    a) ∑(from n=1 to infinity) (arctan n/ n^1.2)
    Comparison test:

    let a_n = \frac{arctan(n)}{n^{1.2}}, and let b_n = \frac{\pi}{n^{1.2}}

    Then a_n \leq b_n (because \lim_{n\to\infty} arctan(n) = \frac {\pi}2)

    And b_n is a p-series with p=1.2, which is greater than 1, therefore \sum b_n converges, and because \sum a_n is monotonic and increasing, and bounded above by \sum b_n we know that \sum a_n converges by the comparison test.

    Quote Originally Posted by Rui View Post
    b) ∑(from n=1 to infinity) ( (n^n) / (2n)! )
    Ratio Test:
    a_n =\frac{n^n}{(2n)!}, and a_{n+1} = \frac{(n+1)^{n+1}}{(2(n+1))!} = \frac{(n+1)^{n+1}}{(2n+1)(2n+2)(2n)!}


    \lim_{n\to\infty} \left| \frac {a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \left| \frac{(n+1)^{n+1}}{(2n+1)(2n+2)(2n)!} * \frac{(2n)!}{n^n}\right|

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

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

    Here multiply numerator and denominator by 1 over n^n to get

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

    Now multiply the numerator and denominator by 1/n to get

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

    It should be relatively clear that this converges to zero, (because the numerator converges to e, and the denominator diverges) but if you need me to explain that more, I can.

    So because \lim_{n\to\infty} \left| \frac {a_{n+1}}{a_n}\right| = 0 < 1 the series \sum_n^\infty \frac{n^n}{(2n)!} is absolutely convergent and therefore convergent.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by angel.white View Post
    Comparison test:

    let a_n = \frac{arctan(n)}{n^{1.2}}, and let b_n = \frac{\pi}{n^{1.2}}

    Then a_n \leq b_n (because \lim_{n\to\infty} arctan(n) = \frac {\pi}2)

    And b_n is a p-series with p=1.2, which is greater than 1, therefore \sum b_n converges, and because \sum a_n is monotonic and increasing, and bounded above by \sum b_n we know that \sum a_n converges by the comparison test.



    Ratio Test:
    a_n =\frac{n^n}{(2n)!}, and a_{n+1} = \frac{(n+1)^{n+1}}{(2(n+1))!} = \frac{(n+1)^{n+1}}{(2n+1)(2n+2)(2n)!}


    \lim_{n\to\infty} \left| \frac {a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \left| \frac{(n+1)^{n+1}}{(2n+1)(2n+2)(2n)!} * \frac{(2n)!}{n^n}\right|

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

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

    Here multiply numerator and denominator by 1 over n^n to get

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

    Now multiply the numerator and denominator by 1/n to get

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

    It should be relatively clear that this converges to zero, (because the numerator converges to e, and the denominator diverges) but if you need me to explain that more, I can.

    So because \lim_{n\to\infty} \left| \frac {a_{n+1}}{a_n}\right| = 0 < 1 the series \sum_n^\infty \frac{n^n}{(2n)!} is absolutely convergent and therefore convergent.
    For the second one also consider

    That there \exists{N}\backepsilon\forall{n>N},\frac{n}{(n)!}>  \frac{n^n}{(2n)!}

    Which implies convergence by the comparison test
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Rui View Post
    Use what test?

    a) ∑(from n=1 to infinity) (arctan n/ n^1.2)
    b) ∑(from n=1 to infinity) ( (n^n) / (2n)! )

    Not sure how to do these...when i do it using ratio test, it turns out really weird..so yea i need help.

    Thanks.
    This is an old one, but for b) consider using the Root Test

    \lim_{n\to\infty}\bigg|\frac{n^n}{(2n)!}\bigg|^{\f  rac{1}{n}}=\lim_{n\to\infty}\bigg|\frac{n^n}{\sqrt  {4\pi{n}}(n^n)^2(e^{-n})^24^n}\bigg|^{\frac{1}{n}}=\lim_{n\to\infty}\bi  gg|\frac{n}{(4\pi{n})^{\frac{1}{2n}}4n^2e^{-2}}\bigg|=0<1

    therefore convergent by root test
    Last edited by Mathstud28; June 2nd 2008 at 12:08 PM.
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