Page 2 of 2 FirstFirst 12
Results 16 to 21 of 21

Math Help - Test the convergence of the series

  1. #16
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    It should be (-1)^{2n}, which equals 1 for all n.
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Is this correct

    Quote Originally Posted by redsoxfan325 View Post
    It should be (-1)^{2n}, which equals 1 for all n.
    I have done something could u please check if it is correct or no

    = \lim_{n \to \infty} \frac{(-1)^{2n}}{(n + 2)\sqrt{n + 1}} . n^{\frac{3}{2}}

    = \lim_{n \to \infty} \frac{(-1)^{2n}}{\sqrt{(n + 2)^2 (n + 1)}} . n^{\frac{3}{2}}

    = \lim_{n \to \infty} \frac{(-1)^{2n}}{\sqrt{(n + 4n + 4)(n + 1)}} . n^{\frac{3}{2}}

    = \lim_{n \to \infty} \frac{(-1)^{2n}}{\sqrt{(n^3 + 5n^2 + 8n + 4)}} . n^{\frac{3}{2}}

    = \lim_{n \to \infty} \frac{(-1)^{2n}}{n^{\frac{3}{2}}\sqrt{( 1 + \frac{5}{3n} + \frac{8}{3n^2} + \frac{4}{3n^3})}} . n^{\frac{3}{2}}

    = \lim_{n \to \infty} \frac{(-1)^{2n}}{ \sqrt{( 1 + \frac{5}{3n} + \frac{8}{3n^2} + \frac{4}{3n^3})}}

    = \frac{1}{\sqrt{1}}

    = 1 Is this Correct ????????
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by zorro View Post
    I have done something could u please check if it is correct or no

    = \lim_{n \to \infty} \frac{(-1)^{2n}}{(n + 2)\sqrt{n + 1}} . n^{\frac{3}{2}}

    = \lim_{n \to \infty} \frac{(-1)^{2n}}{\sqrt{(n + 2)^2 (n + 1)}} . n^{\frac{3}{2}}

    = \lim_{n \to \infty} \frac{(-1)^{2n}}{\sqrt{(n + 4n + 4)(n + 1)}} . n^{\frac{3}{2}}

    = \lim_{n \to \infty} \frac{(-1)^{2n}}{\sqrt{(n^3 + 5n^2 + 8n + 4)}} . n^{\frac{3}{2}}

    = \lim_{n \to \infty} \frac{(-1)^{2n}}{n^{\frac{3}{2}}\sqrt{( 1 + \frac{5}{3n} + \frac{8}{3n^2} + \frac{4}{3n^3})}} . n^{\frac{3}{2}}

    = \lim_{n \to \infty} \frac{(-1)^{2n}}{ \sqrt{( 1 + \frac{5}{3n} + \frac{8}{3n^2} + \frac{4}{3n^3})}}

    = \frac{1}{\sqrt{1}}

    = 1 Is this Correct ????????
    Quote Originally Posted by redsoxfan325 View Post
    So far, yes. Now calculate the limit. You should find that it equals 1, which tells you...
    .
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Is this correct

    Quote Originally Posted by Defunkt View Post
    .

    Since the result is 1 the series convergences ....Am i correct
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Yes, it converges because \sum_{n=1}^{\infty}\frac{1}{n^{3/2}} does.
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Thanks u very much

    Quote Originally Posted by redsoxfan325 View Post
    Yes, it converges because \sum_{n=1}^{\infty}\frac{1}{n^{3/2}} does.


    Thank u redsoxfan325 , Defunkt and Mr fantastic
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Test the series for convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 15th 2010, 04:29 PM
  2. Replies: 4
    Last Post: February 10th 2010, 08:08 AM
  3. Test the convergence of the series
    Posted in the Calculus Forum
    Replies: 9
    Last Post: December 23rd 2009, 02:06 AM
  4. test for convergence of series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 16th 2009, 09:21 AM
  5. Test the Convergence of the series
    Posted in the Calculus Forum
    Replies: 7
    Last Post: December 23rd 2008, 03:39 PM

Search Tags


/mathhelpforum @mathhelpforum