Results 1 to 9 of 9

Math Help - Interval of convergence

  1. #1
    Member
    Joined
    Jan 2011
    Posts
    110

    Interval of convergence

    I need to determine the interval of convergence of the power series

    infinity
    sum of series
    n=1

    (x+2)^n/3n5^n

    I have got as far as

    Here An=1/(3n5^n), for n=1,2,...

    |An+1/An|=1/(3n+1)5^n+1 times 3n5^n/1

    = 1/(3n+1/n)^5 ---> 1/5 as n ----> infinity.

    We have R=5, by the Ratio Test. Since a=?

    Can anyone help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Interval of convergence

    Quote Originally Posted by Arron View Post
    I need to determine the interval of convergence of the power series

    infinity
    sum of series
    n=1

    (x+2)^n/3n5^n

    I have got as far as

    Here An=1/(3n5^n), for n=1,2,...

    |An+1/An|=1/(3n+1)5^n+1 times 3n5^n/1

    = 1/(3n+1/n)^5 ---> 1/5 as n ----> infinity.

    We have R=5, by the Ratio Test. Since a=?

    Can anyone help?
    Shouldn't a_n=\frac{(x+2)^n}{3n*5^n}?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2011
    Posts
    110

    Re: Interval of convergence

    Maybe but does (x+2)^n = 1 in this instance?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,520
    Thanks
    1405

    Re: Interval of convergence

    Quote Originally Posted by Arron View Post
    Maybe but does (x+2)^n = 1 in this instance?
    What would give you that idea?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1584
    Awards
    1

    Re: Interval of convergence

    Quote Originally Posted by Arron View Post
    I need to determine the interval of convergence of the power series
    infinity
    sum of series
    n=1
    (x+2)^n/3n5^n
    Here is what you must do.
    \lim _{n \to \infty } \sqrt[n]{{\frac{{\left| {x + 2} \right|^n}}{{3n \cdot 5^n }}}} \to ? < 1.

    Be sure to test the end points.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Interval of convergence

    If you would do it with the ratio test then you get:
    \lim_{n \to +\infty} \left|\frac{(x+2)^{n+1}}{3(n+1)\cdot 5^{n+1}}\cdot \frac{3n\cdot 5^n}{(x+2)^n}\right|
    =\lim_{n \to +\infty} \left|\frac{(x+2)^{n+1}}{(x+2)^n}\cdot \frac{3n}{3(n+1)}\cdot \frac{5^n}{5^n\cdot 5}\right|
    =\frac{x+2}{5}\lim_{n\to +\infty} \left|\frac{3n}{3n+1}\right|
    =\frac{3(x+2)}{5}

    Now you have two different cases ...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Re: Interval of convergence

    Quote Originally Posted by Siron View Post
    If you would do it with the ratio test then you get:
    \lim_{n \to +\infty} \left|\frac{(x+2)^{n+1}}{3(n+1)\cdot 5^{n+1}}\cdot \frac{3n\cdot 5^n}{(x+2)^n}\right|
    =\lim_{n \to +\infty} \left|\frac{(x+2)^{n+1}}{(x+2)^n}\cdot \frac{3n}{3(n+1)}\cdot \frac{5^n}{5^n\cdot 5}\right|
    =\frac{x+2}{5}\lim_{n\to +\infty} \left|\frac{3n}{3n+1}\right|
    =\frac{3(x+2)}{5}

    Now you have two different cases ...
    \lim_{n \to +\infty} \left|\frac{(x+2)^{n+1}}{3(n+1)\cdot 5^{n+1}}\cdot \frac{3n\cdot 5^n}{(x+2)^n}\right|
    =\lim_{n \to +\infty} \left|\frac{(x+2)^{n+1}}{(x+2)^n}\cdot \frac{3n}{3(n+1)}\cdot \frac{5^n}{5^n\cdot 5}\right|
    =\frac{x+2}{5}\lim_{n\to +\infty} \left|\frac{3n}{3(n+1)}\right|
    =\frac{x+2}{5}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,629
    Thanks
    430

    Re: Interval of convergence

    ratio test to determine an interval of convergence ...

    \lim_{n \to \infty} \left|\frac{(x+2)^{n+1}}{3(n+1)\cdot 5^{n+1}}\cdot \frac{3n\cdot 5^n}{(x+2)^n}\right| < 1

    \lim_{n \to \infty} \left|\frac{(x+2)^{n+1}}{(x+2)^n}\cdot \frac{3n}{3(n+1)}\cdot \frac{5^n}{5^n\cdot 5}\right| < 1

    \left|\frac{x+2}{5}\right| \lim_{n\to \infty} \frac{3n}{3(n+1)} < 1

    \left|\frac{x+2}{5}\right| < 1

    ... don't forget to check the endpoints.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Jan 2011
    Posts
    110

    Re: Interval of convergence

    Hi!

    Thanks for your help everyone. Can we use the Quotient or Squeeze Rule to check for convergence?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. interval of convergence help
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 6th 2011, 04:48 PM
  2. Replies: 1
    Last Post: May 13th 2010, 01:20 PM
  3. Replies: 2
    Last Post: May 1st 2010, 09:22 PM
  4. interval convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 27th 2010, 07:39 PM
  5. convergence interval
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 23rd 2009, 07:14 PM

Search Tags


/mathhelpforum @mathhelpforum