Results 1 to 3 of 3

Math Help - series: integral test

  1. #1
    Member
    Joined
    Nov 2005
    Posts
    172

    series: integral test

    Use the Integral Test to decide the convergence or divergence of the following series:
    \sum_{k=1}^{\infty} \dfrac{k^{2}}{e^{k}}

    I'm pretty much lost with this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    \displaystyle \int_1^{\infty}\frac{x^2}{e^x}dx=\lim_{n\to\infty}  \int_1^nx^2e^{-x}dx=\frac{5}{e} so the serie is convergent.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2007
    Posts
    90
    The integral test is as follows:
    Theorem: The Integral Test
    Let f(x) be a positive continuous function that is eventually decreasing and let f(n)=a_n.
    Then \sum_{n=1}^\infty a_n converges if and only if \int_1^\infty f(x)dx converges and diverges otherwise.
    So to check \sum_{k=1}^{\infty} \dfrac{k^{2}}{e^{k}} for convergence using the integral test, you would find its integral which, like red_dog said, is as follows:

    \int_1^\infty\frac{x^{2}}{e^x}dx=\lim_{b\to\infty}  \int_1^bx^2e^{-x}dx=\lim_{b\to\infty} \left( -x^2e^{-x} - 2xe^{-x}-2e^{-x}\right)\big|_1^b
    =\lim_{b\to\infty} \left( -b^2e^{-b} - 2be^{-b}-2e^{-b}-\left(-1^2e^{-1} - 2e^{-1}-2e^{-1}\right)\right) =0-0-0-\left(-e^{-1}-2e^{-1}-2e^{-1}\right)=\frac{5}{e}

    by using integration by parts twice and then evaluating the limit.

    Therefore, by the Integral Test, because the integral is convergent, so is the series.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Series: Integral test
    Posted in the Calculus Forum
    Replies: 11
    Last Post: September 12th 2009, 05:49 PM
  2. using integral test for sum of a series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 29th 2009, 12:34 PM
  3. Integral test for series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 24th 2009, 08:49 PM
  4. series with integral test
    Posted in the Calculus Forum
    Replies: 13
    Last Post: March 22nd 2009, 12:45 PM
  5. p series + integral test.
    Posted in the Calculus Forum
    Replies: 10
    Last Post: January 14th 2008, 08:00 AM

Search Tags


/mathhelpforum @mathhelpforum