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Thread: 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)
    $\displaystyle
    \forall x\in \mathbb{R},\,\,|\arctan x|<\frac{\pi}{2}$ : you may try the comparison test

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

    Then, you can use $\displaystyle n+1<n$ to say that $\displaystyle (n+1)^{n+1}<n^{n+1}$ 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 $\displaystyle a_n = \frac{arctan(n)}{n^{1.2}}$, and let $\displaystyle b_n = \frac{\pi}{n^{1.2}}$

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

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

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


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

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

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

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

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

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

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

    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 $\displaystyle \lim_{n\to\infty} \left| \frac {a_{n+1}}{a_n}\right| = 0 < 1$ the series $\displaystyle \sum_n^\infty \frac{n^n}{(2n)!}$ 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 $\displaystyle a_n = \frac{arctan(n)}{n^{1.2}}$, and let $\displaystyle b_n = \frac{\pi}{n^{1.2}}$

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

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



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


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

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

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

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

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

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

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

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

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

    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

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

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