Results 1 to 5 of 5

Math Help - Calculus 4: Alternating Series

  1. #1
    Junior Member
    Joined
    May 2007
    Posts
    69

    Calculus 4: Alternating Series

    I have no clue how to solve these. Someone please help...

    1) How to determine the convergence or divergence of the following series:

    a) lim of ((-1)^(n+1) *n)/ (2n-1) when n=1

    b) lim of ((-1)^(n+1) *n^2)/ (n^2 + 5) when n=1

    c) lim of (2(-1)^(n+1))/ (e^n + e^-n)= lim of ((-1)^(n+1) * sech n) when n=1


    2) Determine whether the following series converges conditionally or absolutely, or diverges.

    a) lim of ((-1)^(n+1))/ (n+1) when n=1

    b) lim of ((-1)^n * e^(-n^2)) when n=0

    c) lim of ((-1)^(n+1) * arctan n) when n=1

    3) Using the alternating Series Remainder Theorem, determine the number of terms required to approximate the sum of the following series with an error of less than 0.001.

    a) lim of ((-1)^(n+1)) / (n^2) when n=1
    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
    Use the test for Divergence

    Theorem:

    if \lim_{n\to\infty}a_n\ne0 or does not exits the series

    \sum_{n=1}^\infty a_n is divergent

    This should help with a few. you should look up AST AST estimation theorem.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2007
    Posts
    69
    I don't understand could you please expand on the explanation?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by googoogaga View Post
    I don't understand could you please expand on the explanation?
    What don't you understand? The Theorem? Please be clear.

    \sum_{n=1}^\infty\frac{(-1)^{n+1}n}{2n-1}

    so

    a_n=\frac{(-1)^{n+1}n}{2n-1}

    So divide both the numerator and denominator by n

    a_n=\frac{(-1)^{n+1}}{2-\frac{1}{n}}

    as n goes to infinity the SEQUENCE of a_n diverges.

    so by the Divergence Theorem the SERIES divierges
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    May 2007
    Posts
    69
    Thank you Empty Set. the first part makes sense now that you've broken it down for me. The other ones (2& 3) I am not so sure about...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Alternating series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 24th 2010, 08:50 AM
  2. Alternating series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 5th 2010, 08:43 AM
  3. alternating series #3
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 29th 2009, 06:48 PM
  4. Alternating Series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 18th 2009, 02:26 PM
  5. alternating series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 19th 2008, 03:39 PM

Search Tags


/mathhelpforum @mathhelpforum