Results 1 to 9 of 9

Math Help - radius of convergence

  1. #1
    TS1
    TS1 is offline
    Newbie
    Joined
    Apr 2010
    Posts
    19

    radius of convergence

    i have to determine the radius of convergence of the following power series, state the regions which the series converge uniformly and study the convergence at the boundary of the interval of convergence

    1)

    <br />
\sum_{n=0}^\infty k! x^n<br />

    2)

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

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,569
    Thanks
    1428
    Quote Originally Posted by TS1 View Post
    i have to determine the radius of convergence of the following power series, state the regions which the series converge uniformly and study the convergence at the boundary of the interval of convergence

    1)

    <br />
\sum_{n=0}^\infty k! x^n<br />

    2)

    <br />
\sum_{n=1}^\infty \frac{(n!)^2}{(2n)!}x^n<br />
    2. Use the ratio test.

    \left|\frac{a_{n + 1}}{a_n}\right| = \left|\frac{[(n + 1)!]^2 (2n)!x^{n + 1}}{(n!)^2 x^n [2(n + 1)]!}\right|

     = \left|\frac{(n + 1) x}{2(2n + 1)}\right|.


    So \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty}\left| \frac{(n + 1) x}{2(2n + 1)}\right|

     = \lim_{n \to \infty}\left[ \left(\frac{n + 1}{2n + 1}\right) \left|\frac{x}{2}\right|\right]

     = \lim_{n \to \infty}\left[\left(\frac{1}{2}\right)\left|\frac{x}{2}\right|\r  ight] by L'Hospital's Rule

     = \left|\frac{x}{4}\right|.


    For this series to converge, the limit is <1.

    So \left|\frac{x}{4}\right|<1

    -1 <\frac{x}{4} < 1

    -4 < x < 4.


    So the radius of convergence is 4. I'll leave you to check the end points.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    TS1
    TS1 is offline
    Newbie
    Joined
    Apr 2010
    Posts
    19
    what happens to the convergence at the boundary of the interval of convergence?
    thank you
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,714
    Thanks
    1472
    Quote Originally Posted by TS1 View Post
    i have to determine the radius of convergence of the following power series, state the regions which the series converge uniformly and study the convergence at the boundary of the interval of convergence

    1)

    <br />
\sum_{n=0}^\infty k! x^n<br />
    Is that really k! and not n!? If so, the sum can be written as k!\sum_{n=0}^\infty x^n and the radius of convergence is just the radius of convergence of \sum_{n=0}^\infty x^n. As Prove It says, the simplest way to find radii of convergence for power series is to use the ratio test:
    If a_n= x^n, then a_{n+1}= x^{n+1} and \left|\frac{a_{n+1}}{a_n}\right|= \left|\frac{x^{n+1}}{x^n}\right|= |x|. Since the ratio test says that will converge uniformly for the limit of the ratio less than one, we must have |x|< 1 and the radius of convergence is 1.
    Now, the boundary points are x= -1 and x= 1. If x= 1, that is k(1+ 1+ 1+ ...). Does that converge? At x= -1, it is k(1- 1+ 1- 1+ ....). Does that converge?

    If it is actually \sum_{n=0}^\infty n! x^n it's a little different:
    \left|\frac{a_{n+1}}{a_n}\right|= \left|\frac{(n+1)x^{n+1}}{nx^n}\right|= \frac{n+1}{n} |x| and you must take the limit, as n goes to infinity, of that.

    2)

    <br />
\sum_{n=1}^\infty \frac{(n!)^2}{(2n)!}x^n<br />
    Last edited by HallsofIvy; April 17th 2010 at 08:07 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,569
    Thanks
    1428
    Quote Originally Posted by TS1 View Post
    what happens to the convergence at the boundary of the interval of convergence?
    thank you
    Since the ratio test fails when the limit is 1, you need to use some other test at the boundaries (i.e. where the limit is 1).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    TS1
    TS1 is offline
    Newbie
    Joined
    Apr 2010
    Posts
    19
    for the second question i should use a different test?? i'm stuck on what to do and what my answer implies?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,714
    Thanks
    1472
    Post it has already told you that the radius of convergence is 4. That means that the endpoints of the interval of convergence are x= -4 and x= 4.

    For x= 4, that series is \sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}4^n or 1+ 2+ \frac{4}{3}+ \frac{24}{5}+ \cdot\cdot\cdot
    That's not going to be easy to decide!

    For x= -4 that series is \sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}(-4)^n. Because that is an 'alternating' series, it will converge as long as the individual terms are eventually decreasing.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,569
    Thanks
    1428
    Quote Originally Posted by HallsofIvy View Post
    Post it has already told you that the radius of convergence is 4. That means that the endpoints of the interval of convergence are x= -4 and x= 4.

    For x= 4, that series is \sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}4^n or 1+ 2+ \frac{4}{3}+ \frac{24}{5}+ \cdot\cdot\cdot
    That's not going to be easy to decide!

    For x= -4 that series is \sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}(-4)^n. Because that is an 'alternating' series, it will converge as long as the individual terms are eventually decreasing.
    Use the simple divergence test, that if \lim_{n \to \infty}a_n \neq 0 then the series is divergent...

    In this case, you have

    \sum_{n = 1}^{\infty}\frac{(n!)^24^n}{(2n)!}

    According to Wolfram, this limit is \infty, not 0.

    So the series is divergent when x = 4.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    TS1
    TS1 is offline
    Newbie
    Joined
    Apr 2010
    Posts
    19
    thanks, but how would i find the limit a(n)?
    <br />
    " alt="\sum_{n = 1}^{\infty}\frac{(n!)^24^n}{(2n)!}
    " />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: March 27th 2011, 07:42 PM
  2. Replies: 1
    Last Post: May 13th 2010, 01:20 PM
  3. Replies: 2
    Last Post: May 1st 2010, 09:22 PM
  4. Replies: 1
    Last Post: November 13th 2009, 06:42 AM
  5. series convergence and radius of convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 15th 2008, 08:07 AM

Search Tags


/mathhelpforum @mathhelpforum