Results 1 to 12 of 12

Math Help - Convergence radius and interval

  1. #1
    Junior Member Revy's Avatar
    Joined
    Mar 2010
    From
    Utena/Vilnius
    Posts
    39

    Convergence radius and interval

    I need to find radius of convergence and interval of convergence of \sum_{n=1}^{\inf}\frac{x^n}{\sqrt{n}}

    In d'Alembert feature I find radius like this \lim_{n->\inf}\frac{x^{n+1}\sqrt{n}}{\sqrt{n+z}x^n}=|x|<  1
    |x|<1
    -1<x<1

    Radius is 1

    If I try in Cauchy feature, I get same:
    \lim_{n->\inf}\sqrt{|\frac{x^n}{\sqrt{n}}|}<1 first square n root
    \lim_{n->\inf}|\frac{x}{\sqrt_{n+2}{n}}}|<1 square n+2 root
    |x|<1
    -1<x<1

    Radius is 1



    Interval of convergence:
    I take -1 and 1
    \sum_{n=1}^{\inf}\frac{x^2 * 1^n}{\sqrt{n}} and \sum_{n=1}^{\inf}\frac{x^2 * (-1)^n}{\sqrt{n}}

    I do check if diverge/converge:
    \lim_{n->\inf}\frac{x^n}{\sqrt{n}}=\inf it diverge
    \lim_{n->\inf}|\frac{x^n}{\sqrt{n}}|!=0 here != means 'not even'

    Interval is x\E(-1;1)

    Questions:
    1. Did I do task correct?
    2. \lim_{n->\inf}|\frac{x^n}{\sqrt{n}}|!=0 it diverge or converge?
    3. How interval of convergence change because of divergence and convergence?
    Last edited by Revy; August 24th 2010 at 01:53 PM. Reason: gramma correction
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    You need to ask if \displaystyle\sum\limits_n {\frac{{( - 1)^n }}<br />
{{\sqrt n }}} converges.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member Revy's Avatar
    Joined
    Mar 2010
    From
    Utena/Vilnius
    Posts
    39
    Quote Originally Posted by Plato View Post
    You need to ask if \displaystyle\sum\limits_n {\frac{{( - 1)^n }}<br />
{{\sqrt n }}} converges.
    It doesn't have limit, but it still get two answers: 1 and -1. If it's like this, it is still said that it converges?

    How does answer change because of that?
    How would answer change if it converged with limit 1/2?
    And how would answer change if it would converge with limit 2?
    Last edited by Revy; August 26th 2010 at 01:52 AM. Reason: gramma
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,315
    Thanks
    1225
    a_{n + 1} = \frac{x^{n + 1}}{\sqrt{n + 1}} and a_n = \frac{x^n}{\sqrt{n}}.


    By the ratio test, the series converges where

    \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right| < 1

    \lim_{n \to \infty}\left|\frac{\frac{x^{n + 1}}{\sqrt{n + 1}}}{\frac{x^n}{\sqrt{n}}}\right| < 1

    \lim_{n \to \infty}\left|\frac{x\sqrt{n}}{\sqrt{n + 1}}\right| < 1

    \lim_{n \to \infty}\sqrt{\frac{n}{n + 1}}|x| < 1

    \lim_{n \to \infty}\sqrt{\frac{n + 1 - 1}{n + 1}}|x| < 1

    \lim_{n \to \infty}\sqrt{1 - \frac{1}{n + 1}}|x| < 1

    1|x| < 1

    |x| < 1.


    So the radius of convergence is 1 and the interval of convergence is -1 < x < 1.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member Revy's Avatar
    Joined
    Mar 2010
    From
    Utena/Vilnius
    Posts
    39
    Quote Originally Posted by Prove It View Post
    So the radius of convergence is 1 and the interval of convergence is -1 < x < 1.
    Isn't interval (-1;1) only when it diverge with -1 and 1?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,315
    Thanks
    1225
    By the ratio test, the series converges where the limit < 1, diverges where the limit is > 1 and inconclsive where the limit  = 1.

    So all we can say at the moment is that the series is convergent when |x| < 1, because this is where the limit < 1.


    You will need to use a different test to test the endpoints.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,306
    Thanks
    1282
    Specifically, when x= 1, the series becomes \sum \frac{1}{\sqrt{n}}= \sum \frac{1}{x^{1/2}} and 1/2< 1. What does that tell you?

    When x= -1, the series becomes \sum\frac{(-1)^n}{\sqrt{n}} which is an alternating series and \frac{1}{\sqrt{n}} decreases to 0 as n goes to infinity. What does that tell you?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member Revy's Avatar
    Joined
    Mar 2010
    From
    Utena/Vilnius
    Posts
    39
    Both of them converge.
    Is convergence and divergence important finding interval of convergence? If so, how?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Are you sure \displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} converges?

    And what do you mean by that question? Convergence and divergence of what?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,315
    Thanks
    1225
    Quote Originally Posted by Revy View Post
    Both of them converge.
    Is convergence and divergence important finding interval of convergence? If so, how?
    Well of course it is. You're testing if the function will converge at the endpoints. So obviously you need to see if the endpoints converge or diverge.


    Specifically for the case of x = 1 when the series becomes \sum{\frac{1}{\sqrt{n}}} = \sum{\frac{1}{n^{\frac{1}{2}}}}, this is a p-series with p < 1. What do you know about p-series?


    For the case of x = -1 when the series becomes \sum{\frac{(-1)^n}{\sqrt{n}}}, what are the conditions for which Leibnitz's Alternating Series Theorem implies convergence?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member Revy's Avatar
    Joined
    Mar 2010
    From
    Utena/Vilnius
    Posts
    39
    Quote Originally Posted by Prove It View Post
    What do you know about p-series?
    If p<=1, then series diverge.
    Quote Originally Posted by Prove It View Post
    What are the conditions for which Leibnitz's Alternating Series Theorem implies convergence?
    If sequence monotonically decrease to 0, then series converge.
    Last edited by Revy; August 27th 2010 at 01:26 PM.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,315
    Thanks
    1225
    Correct, so putting it all together you know that the series converges for -1 < x < 1, diverges for x < -1 and x > 1, diverges at x = 1 and converges at x = -1.

    So the interval of convergence is

    -1 \leq x < 1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Radius and Interval of Convergence
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 2nd 2010, 08:57 AM
  2. Replies: 2
    Last Post: May 1st 2010, 09:22 PM
  3. radius and interval of convergence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 7th 2009, 12:44 PM
  4. Radius and interval of convergence?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 30th 2009, 08:08 PM
  5. Radius and interval of convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 25th 2009, 01:30 AM

Search Tags


/mathhelpforum @mathhelpforum