Results 1 to 4 of 4

Math Help - convergence

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

    convergence

    1st) I need to investigate convergence using convergence of integral feature
    \sum_{n=1}^{inf}\frac{1}{\sqrt{n}}
    From limit I see that it's 0, but how do I prove it using that method?


    2nd) Here I need to investigate convergence using series required condition of convergence
    \sum_{n=1}^{\inf}\cos^2n
    Last edited by Revy; August 27th 2010 at 01:02 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    The series...

    \displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} (1)

    ... diverges and You can prove that considering that is...

    \displaystyle \sum_{n=1}^{N} \frac{1}{\sqrt{n}} > 1 + \int_{1}^{N} \frac{dx}{\sqrt{1+x}} (2)

    But a primitive of the function under integral sign in (2) is \displaystyle  \frac{\sqrt{1+x}}{2} so that the integral in (2) diverges and the same is for the series (1)...

    Kind regards

    \chi \sigma
    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 chisigma View Post
    But a primitive of the function under integral sign in (2) is \displaystyle  \frac{\sqrt{1+x}}{2}
    isn't primitive 2\sqrt{x+1}+C ?
    this one also diverge so proving doesn't change. Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2010
    Posts
    37
    for the 1)st question you know that 1/ \sqrt{n} \ge 1/n \forall n \in N (natural numbers). and you know that \sum_{n=1}^{\infty}\frac{1}{n} is the harmonic series that diverges. so by comparison test \sum_{n=1}^{\infty}\frac{1}\sqrt{n} diverges. since \sum1/n diverges \forall c \not= 0 c \sum_{n=1}^{\infty}1/n diverges. therefore the radius of convergence is 0.

    2nd) there is a theorem that states that \sum_{n=1}^{\infty}a_n converges \Longrightarrow \lim_{n\to\infty}a_n= 0
    which means if \lim_{n\to\infty}a_n\not=0 \Longrightarrow \sum_{n=1}^{\infty}a_n diverges. since 1 \ge \cos^2n \ge0 \forall n \in N, for a given n you can find n \prime\in N such that cos n \prime\not=0, which means \cos^2n \prime\not=0. therefore \sum_{n=1}^{\infty}\cos^2{n} diverges. once again the radius of convergence is 0.
    hope this is helpful.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. dominated convergence theorem for convergence in measure
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: October 1st 2014, 12:11 PM
  2. Replies: 7
    Last Post: March 27th 2011, 07:42 PM
  3. Replies: 1
    Last Post: May 13th 2010, 01:20 PM
  4. Replies: 2
    Last Post: May 1st 2010, 09:22 PM
  5. Replies: 6
    Last Post: October 1st 2009, 09:10 AM

/mathhelpforum @mathhelpforum