Results 1 to 10 of 10

Math Help - Series

  1. #1
    Member
    Joined
    Nov 2009
    Posts
    81

    Question Series

    Hi

    I am find it hard to figure this series out?

    Would anyone have any ideas?

    1 +2/3+ 8/15 + 48/105 +384/945+ 3840/10395 + 46080/135135 + 645120/2027025 +10321920/34459425+...
    Last edited by hebby; February 9th 2010 at 06:22 PM. Reason: Incorrect numbers
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by hebby View Post
    Hi

    I am find it hard to figure this series out?

    Would anyone have any ideas?

    1 +2+ 8 + 48 +384+ 3840 + 46080 + 645120 +10321920+...
    3 15 105 945 10395 135135 2027025 34459425
    This is

    T_1=1

    T_{n+1}=T_n(2n)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    81

    Unhappy

    sorry the numbers were incorrect
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by hebby View Post
    Hi

    I am find it hard to figure this series out?

    Would anyone have any ideas?

    1 +2/3+ 8/15 + 48/105 +384/945+ 3840/10395 + 46080/135135 + 645120/2027025 +10321920/34459425+...
    This is T_1=1

    T_{n+1}=T_n\left(\frac{2n}{2n+1}\right)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2009
    Posts
    81

    Question

    hi

    then how would you write it in the sigma form?

    I have to use the ratio test
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Hi hebby,

    there is no need to use the sigma format in this case.

    Instead.

    a_{n+1}=a_n\left(\frac{2n}{2n+1}\right)

    \frac{a_{n+1}}{a_n}=\frac{2n}{2n+1}

    As n approaches infinity, the numerator and denominator become indistinguishable, hence the limit of this ratio as n approaches infinity is 1.
    The ratio test is inconclusive for this series, it does not tell you if the series converges or diverges.

    \lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=\li  m_{n\rightarrow\infty}|\frac{T_{n+1}}{T_n}|=1
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Nov 2009
    Posts
    81

    Question

    Hi

    When I apply the Raabe Test,

    Rn= lim (n(1-(an+1/an)), I get 1/2, does that mean it diverges?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member Renji Rodrigo's Avatar
    Joined
    Sep 2009
    From
    Rio de janeiro
    Posts
    38
    the partial sum is \sum^{n}_{k=0}\frac{4^{k}}{ (2k+1){2k \choose k} }=\frac{4^{n+1}}{ {2n+2 \choose n+1} }-1.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by hebby View Post
    Hi

    When I apply the Raabe Test,

    Rn= lim (n(1-(an+1/an)), I get 1/2, does that mean it diverges?
    Yes, hebby.

    There is no question of conditional convergence, since it's not an alternating series.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by hebby View Post
    Hi

    I am find it hard to figure this series out?

    Would anyone have any ideas?

    1 +2/3+ 8/15 + 48/105 +384/945+ 3840/10395 + 46080/135135 + 645120/2027025 +10321920/34459425+...
    Show by induction that T_n\geqslant\frac{1}{n}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 3rd 2011, 01:12 AM
  2. Replies: 3
    Last Post: September 29th 2010, 06:11 AM
  3. Replies: 0
    Last Post: January 26th 2010, 08:06 AM
  4. Replies: 2
    Last Post: September 16th 2009, 07:56 AM
  5. Replies: 1
    Last Post: May 5th 2008, 09:44 PM

Search Tags


/mathhelpforum @mathhelpforum