Results 1 to 5 of 5

Math Help - Testing a series for convergence or divergence

  1. #1
    Newbie
    Joined
    Apr 2010
    From
    Home of country music
    Posts
    22

    Testing a series for convergence or divergence

    I am trying to test this series by the alternating series test:

    \sum\frac{\sin{n}\frac\pi2}{n!}

    n=1 to infinity

    I don't really know how to find the limit or how to see if it is decreasing, any help would be greatly appreciated.

    Thanks,
    CC
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by CalculusCrazed View Post
    I am trying to test this series by the alternating series test:

    \sum\frac{\sin{n}\frac\pi2}{n!}

    n=1 to infinity

    I don't really know how to find the limit or how to see if it is decreasing, any help would be greatly appreciated.

    Thanks,
    CC
    \sin\left(\frac{n\pi}{2}\right)=(-1)^n
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2010
    From
    Home of country music
    Posts
    22
    So I basically have:

    \lim\frac{1}{n!}
    as n goes to infinity

    which would equal zero

    and decreasing because the first term is
    1/1 then the second is 1/(2*1) and so on?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by CalculusCrazed View Post
    So I basically have:

    \lim\frac{1}{n!}
    as n goes to infinity

    which would equal zero

    and decreasing because the first term is
    1/1 then the second is 1/(2*1) and so on?
    Oh wow, I didn't even notice that it was n!.

    With my hint you can use the alternating series test.

    You can also note that \left|\sum_{n=1}^{\infty}\frac{\sin\left(\frac{\pi n}{2}\right)}{n!}\right|\leqslant \sum_{n=1}^{\infty}\frac{\left|\sin\left(\tfrac{\p  i n}{2}\right)\right|}{n!} \leqslant\sum_{n=1}^{\infty}\frac{1}{n!}=e

    Or the ratio test even works nice.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2010
    From
    Home of country music
    Posts
    22
    Ahh, okay cool. Thanks a ton!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Testing Series for Convergence/Divergence
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 16th 2010, 04:39 PM
  2. Testing series for convergence or divergence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 27th 2010, 09:33 PM
  3. testing for convergence/divergence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 23rd 2010, 05:44 AM
  4. Testing for Convergence/Divergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 26th 2009, 07:19 PM
  5. Replies: 3
    Last Post: October 8th 2009, 10:24 AM

Search Tags


/mathhelpforum @mathhelpforum