Results 1 to 3 of 3

Math Help - Interval of Convergence Problem

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    13

    Post Interval of Convergence Problem

    Sorry, but I have no idea how to use the math symbol thingies, so bear with me here.

    How would I go about finding the interval of convergence of:

    Σ (((-1)^n)(x^n))/n
    Also, can the endpoints be included in the interval?
    Thanks so much.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,796
    Thanks
    1576
    Quote Originally Posted by suchgreatheights View Post
    Sorry, but I have no idea how to use the math symbol thingies, so bear with me here.

    How would I go about finding the interval of convergence of:

    Σ (((-1)^n)(x^n))/n
    Also, can the endpoints be included in the interval?
    Thanks so much.
    You will need to use the Ratio Test.

    Remembering that the Ratio Test tells us that the series is convergent where \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right|<1, divergent where \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right|>1 and inconclusive where \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right|=1, no, you can not use the endpoints in your interval. However, there are other tests you can do for the endpoints.


    Anyway, you have \sum_{n = 0}^{\infty}\frac{(-1)^nx^n}{n}.

    This will be convergent where

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

    \lim_{n \to \infty}\left|\frac{\frac{(-1)^{n + 1}x^{n + 1}}{n + 1}}{\frac{(-1)^nx^n}{n}}\right| < 1

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

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

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

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

    |x| \lim_{n \to \infty}\left(1 - \frac{1}{n + 1}\right) < 1

    |x| (1) < 1

    |x| < 1

    -1 < x < 1.


    So the series is convergent where -1 < x < 1 and divergent where x < -1 and x > 1. You will need to use a different test for the endpoints of the interval.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,056
    Thanks
    894
    Quote Originally Posted by suchgreatheights View Post
    Sorry, but I have no idea how to use the math symbol thingies, so bear with me here.

    How would I go about finding the interval of convergence of:

    Σ (((-1)^n)(x^n))/n
    Also, can the endpoints be included in the interval?
    Thanks so much.
    set up the ratio test ...

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

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

    |x| \cdot 1 < 1

    -1 < x < 1

    if x = -1 , nth term is \frac{1}{n} ... the harmonic series diverges.

    if x = 1 , nth term is \frac{(-1)^n}{n} ... alternating harmonic series converges

    interval of convergence is -1 < x \le 1
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Interval of convergence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 25th 2010, 05:08 AM
  2. Replies: 1
    Last Post: May 13th 2010, 02:20 PM
  3. Replies: 2
    Last Post: May 1st 2010, 10:22 PM
  4. Interval of convergence - confusing problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 6th 2009, 12:49 PM
  5. Interval of Convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 10th 2008, 02:49 PM

Search Tags


/mathhelpforum @mathhelpforum