Results 1 to 13 of 13

Math Help - Convergence Problem

  1. #1
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523

    Convergence Problem

    This is the equation that i have ended up with

    \sum_{n=0}^{\infty} \frac{x^{2n}}{(n+2) \sqrt{n+1}}

    I am using the Ratio test \frac{u_{n+1} }{u_n} = k ....if k<1 (convergent) k>1(divergent) ....

    \lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \lim_{n \to \infty} \frac{x^2}{n+3 \sqrt{n+2}} . \frac{n+2 \sqrt{n+1}}{1}

    = x^2 \frac{n+2 \sqrt{n+1}}{n+3 \sqrt{n+2}} .........I am stuck here what should i do next?????
    Last edited by zorro; December 27th 2009 at 01:14 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Your constants are off and you need a ().
    Then let n go to infinity and the limit will be x^2
    set that less than 1 and you will see that in the interval of convergence is (-1,1).
    THEN test the endpoint and since the series will converge at both endpoints, the interval of convergence is [-1,1].
    PLUS you should rewrite the n's as (n+1)^{3/2}
    And this is just calc 2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523
    Quote Originally Posted by matheagle View Post
    Your constants are off and you need a ().
    Then let n go to infinity and the limit will be x^2
    set that less than 1 and you will see that in the interval of convergence is (-1,1).
    THEN test the endpoint and since the series will converge at both endpoints, the interval of convergence is [-1,1].
    PLUS you should rewrite the n's as (n+1)^{3/2}
    And this is just calc 2.
    I didnt understand u mite .........seriously i dont have a clue what ur are telling me

    First the constant in the eq is x^2

    Second how should i get rid of the other terms.....
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2009
    Posts
    263

    Smile

    Quote Originally Posted by zorro View Post
    This is the equation that i have ended up with

    \sum_{n=0}^{\infty} \frac{x^{2n}}{(n+1) \sqrt{n+1}}

    I am using the Ratio test \frac{u_{n+1} }{u_n} = k ....if k<1 (convergent) k>1(divergent) ....

    \lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \lim_{n \to \infty} \frac{x^2}{n+3 \sqrt{n+2}} . \frac{n+2 \sqrt{n+1}}{1}

    = x^2 \frac{n+2 \sqrt{n+1}}{n+3 \sqrt{n+2}} .........I am stuck here what should i do next?????
    hi zorro,
    check your work again,
    what is u_{n}?, u_{n+1}?

    Spoiler:

    \lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \lim_{n \to \infty} \frac{x^2}{(n+2) (\sqrt{n+2})} \times \frac{(n+1) (\sqrt{n+1})}{1} =\lim_{n \to \infty} \frac{x^2}{(n+2)^{3/2}} \times \frac{(n+1)^{3/2}}{1} = x^2
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    and rewrite this series as...

    \sum_{n=0}^{\infty} \frac{x^{2n}}{(n+1)^{3/2}}
    Follow Math Help Forum on Facebook and Google+

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

    Sorry i made a mistake in the question which i posted

    Quote Originally Posted by dedust View Post
    hi zorro,
    check your work again,
    what is u_{n}?, u_{n+1}?

    Spoiler:

    \lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \lim_{n \to \infty} \frac{x^2}{(n+2) (\sqrt{n+2})} \times \frac{(n+1) (\sqrt{n+1})}{1} =\lim_{n \to \infty} \frac{x^2}{(n+2)^{3/2}} \times \frac{(n+1)^{3/2}}{1} = x^2

    Sorry I posted the question falsely the question is this

     \sum_{n=0}^{\infty} \frac{x^{2n}}{(n+2) \sqrt{n+1}}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    It still doesn't matter, the answer is as I stated earlier -1\le x\le 1
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523
    Quote Originally Posted by matheagle View Post
    It still doesn't matter, the answer is as I stated earlier -1\le x\le 1
    But how would u do it ...........i am not able to get x^2
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Nov 2009
    Posts
    263

    Thumbs up

    Quote Originally Posted by zorro View Post
    Sorry I posted the question falsely the question is this

     \sum_{n=0}^{\infty} \frac{x^{2n}}{(n+2) \sqrt{n+1}}
    \lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \lim_{n \to \infty} \frac{x^2}{(n+3) (\sqrt{n+2})} \times \frac{(n+2) (\sqrt{n+1})}{1} =\lim_{n \to \infty} \frac{x^2}{(n\sqrt{n+2}+3\sqrt{n+2})} \times \frac{(n\sqrt{n+1}+2\sqrt{n+1})}{1} = \lim_{n \to \infty} \frac{x^2}{(\sqrt{n^3+2n^2}+3\sqrt{n+2})} \times \frac{(\sqrt{n^3+n^2}+2\sqrt{n+1})}{1} = \lim_{n \to \infty} \frac{x^2}{(\sqrt{n^3+2n^2}+3\sqrt{n+2})} \times \frac{(\sqrt{n^3+n^2}+2\sqrt{n+1})}{1} \times \frac{\frac{1}{n^{3/2}}}{\frac{1}{n^{3/2}}} = \lim_{n \to \infty} \frac{x^2}{(\sqrt{1+2n^{-1}}+3\sqrt{n^{-2}+2n^{-2}})} \times \frac{(\sqrt{1+n^{-1}}+2\sqrt{n^{-2} +n^{-3}})}{1} = x^2

    it will converge if . . .
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,454
    Thanks
    1868
     \frac{u_{n+1}}{u_n}= \frac{x^{2n+2}}{(n+3)\sqrt{n+3}}\frac{(n+2)\sqrt{n  +1}}{x^{2n}}= \frac{x^{2n+2}}{x^{2n}}\frac{n+2}{n+3}\frac{\sqrt{  n+1}}{\sqrt{n+2}}
    = x^2\frac{n+2}{n+3}\sqrt{\frac{n+1}{n+2}}

    Is that clearer?

    Take the limit, as n goes to infinity, by dividing numerator and denominator of each fraction by n.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523
    Quote Originally Posted by dedust View Post
    \lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \lim_{n \to \infty} \frac{x^2}{(n+3) (\sqrt{n+2})} \times \frac{(n+2) (\sqrt{n+1})}{1} =\lim_{n \to \infty} \frac{x^2}{(n\sqrt{n+2}+3\sqrt{n+2})} \times \frac{(n\sqrt{n+1}+2\sqrt{n+1})}{1} = \lim_{n \to \infty} \frac{x^2}{(\sqrt{n^3+2n^2}+3\sqrt{n+2})} \times \frac{(\sqrt{n^3+n^2}+2\sqrt{n+1})}{1} = \lim_{n \to \infty} \frac{x^2}{(\sqrt{n^3+2n^2}+3\sqrt{n+2})} \times \frac{(\sqrt{n^3+n^2}+2\sqrt{n+1})}{1} \times \frac{\frac{1}{n^{3/2}}}{\frac{1}{n^{3/2}}} = \lim_{n \to \infty} \frac{x^2}{(\sqrt{1+2n^{-1}}+3\sqrt{n^{-2}+2n^{-2}})} \times \frac{(\sqrt{1+n^{-1}}+2\sqrt{n^{-2} +n^{-3}})}{1} = x^2

    it will converge if . . .
    How do u know it will converge as x^2 is neither >1 or <1 .........
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member
    Joined
    Nov 2009
    Posts
    263

    Smile

    Quote Originally Posted by zorro View Post

    I am using the Ratio test \lim_{n \to \infty}\frac{u_{n+1} }{u_n} = k ....if k<1 (convergent) k>1(divergent) ....

    here, \lim_{n \to \infty}\frac{u_{n+1} }{u_n} = x^2
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Banned
    Joined
    Dec 2008
    From
    Mauritius
    Posts
    523
    Quote Originally Posted by dedust View Post
    here, \lim_{n \to \infty}\frac{u_{n+1} }{u_n} = x^2
    since the test couldnt find if the series is divergent or not i am now going to use Raabe Ratio test

    \lim_{n \to \infty} n \left(\frac{u_n}{u_{n+1}}-1 \right) = \lim_{n \to \infty} n \left(\frac{1}{x^2} -1 \right) =

    \lim_{n \to \infty} n \left(\frac{1 - x^2}{x^2} \right) = \lim_{n \to \infty}  \left(\frac{n -n. x^2}{n.x^2} \right)

    dividing by 1/n^2 weget

    = 0

    \lim_{n \to \infty} n \left(\frac{u_n}{u_{n+1}}-1 \right) = 0<1 \therefore divergent
    Last edited by zorro; December 27th 2009 at 10:11 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. new convergence problem
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 24th 2010, 08:44 AM
  2. Convergence problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 8th 2010, 07:09 AM
  3. problem about almost sure and L1 convergence
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: November 11th 2009, 12:40 PM
  4. Convergence problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 15th 2009, 11:22 PM
  5. Convergence problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 22nd 2007, 08:00 PM

Search Tags


/mathhelpforum @mathhelpforum