Results 1 to 3 of 3

Thread: Series questions!

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    12

    Series questions!

    (3) / n sqrt(n)

    not sure what test i should use for this problem!


    -----------------------------------------------------

    [( -1 )^n] * [3^(n-2)] / 2^n

    i dont understand why i cannot use this root test for this problem...it seems to work out fine and i get the limit = 0 so since L < 1 then it should be convergent....but this is actually supposed to be divergent.



    ----------------------------------------------------------

    10n+3 / n2^n

    another one i'm not sure which test i should use
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by lochnessmonster View Post
    (3) / n sqrt(n)

    not sure what test i should use for this problem!


    -----------------------------------------------------

    [( -1 )^n] * [3^(n-2)] / 2^n

    i dont understand why i cannot use this root test for this problem...it seems to work out fine and i get the limit = 0 so since L < 1 then it should be convergent....but this is actually supposed to be divergent.



    ----------------------------------------------------------

    10n+3 / n2^n

    another one i'm not sure which test i should use
    I am having a hard time reading the first one my guess is that it is this

    $\displaystyle \displaystyle \sum_{n=0}^{\infty}\frac{3}{n\sqrt{n}}=\sum_{n=0}^ {\infty}\frac{3}{n^{\frac{3}{2}}}$

    This is a p-series with $\displaystyle p=\frac{3}{2}$

    $\displaystyle \displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n3^{n-2}}{2^n}=3^{-2}\sum_{n=0}^{\infty}\left( \frac{-3}{2} \right)^n$

    This is a geometric series or use the basic test for divergence!

    For the last one I think this is the series

    $\displaystyle \displaystyle \sum_{n=0}^{\infty}\frac{10n+3}{n2^n}$

    Use the limit comparison test with $\displaystyle \displaystle b_n=\frac{1}{2^n}$ this is a convergent geometric series
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46
    Quote Originally Posted by lochnessmonster View Post
    (3) / n sqrt(n)

    Use the integral test.


    [( -1 )^n] * [3^(n-2)] / 2^n

    i dont understand why i cannot use this root test for this problem...it seems to work out fine and i get the limit = 0 so since L < 1 then it should be convergent....but this is actually supposed to be divergent.

    Check your limit (it is not $\displaystyle 0$) . The limit $\displaystyle \lim_{n\to +\infty}( -1 )^n 3^{n-2} / 2^n$ does not exist, so the series is divergent.


    10n+3 / n2^n another one i'm not sure which test i should use

    Use the ratio test.


    Edited: Sorry, I didn`t see the previous post.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Series Questions
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Jan 13th 2012, 05:59 AM
  2. Power Series&Function Series Questions
    Posted in the Calculus Forum
    Replies: 10
    Last Post: Jan 19th 2010, 11:38 PM
  3. A series of questions...(2)
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Aug 14th 2009, 12:38 AM
  4. A series of questions...(5)
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: Aug 13th 2009, 04:43 PM
  5. Series questions
    Posted in the Calculus Forum
    Replies: 26
    Last Post: Jun 2nd 2008, 02:23 PM

Search Tags


/mathhelpforum @mathhelpforum