Results 1 to 2 of 2

Thread: Convergence of a Series (proof)

  1. #1
    Senior Member roninpro's Avatar
    Joined
    Nov 2009
    Posts
    485

    Convergence of a Series (proof)

    Hello. I am having some difficulty proving the following:

    If $\displaystyle \sum_{n=1}^\infty a_n$ converges and $\displaystyle \{b_n\}$ is monotonic and bounded, then $\displaystyle \sum_{n=1}^\infty a_nb_n$ converges.

    I tried using the Cauchy Criterion and partial summation to handle this, but I haven't had any luck. Your insight would be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by roninpro View Post
    Hello. I am having some difficulty proving the following:

    If $\displaystyle \sum_{n=1}^\infty a_n$ converges and $\displaystyle \{b_n\}$ is monotonic and bounded, then $\displaystyle \sum_{n=1}^\infty a_nb_n$ converges.

    I tried using the Cauchy Criterion and partial summation to handle this, but I haven't had any luck. Your insight would be appreciated.

    It's a result of Dirichlet's test: if $\displaystyle \sum\limits_{n=1}^\infty a_n$ is a bounded series (i.e., its partials sums sequence is bounded) and if $\displaystyle \{b_n\}_{n=1}^\infty$ is a descending monotone sequence that converges to zero, then $\displaystyle \sum\limits_{n=1}^\infty a_nb_n$ converges:

    Since $\displaystyle \{b_n\}$ is monotone and bounded it converges to a finite limit, say L. Assume it is monotone ascending (if it is descending it is very simmlilar) , so we get $\displaystyle (L-b_n) \xrightarrow [n\to \infty] {} 0$ monotonically descending and $\displaystyle \sum\limits_{n=1}^\infty a_n$ is bounded because it is convergent, and thus by Dirichlet's test the series $\displaystyle \sum\limits_{n=1}^\infty a_n(L-b_n)$ converges, but then:

    $\displaystyle \sum\limits_{n=1}^\infty a_nL-\sum\limits_{n=1}^\infty a_n(L-b_n)$ is the difference of two convergent series and thus it converges, and $\displaystyle \sum\limits_{n=1}^\infty a_nL-\sum\limits_{n=1}^\infty a_n(L-b_n)=\sum\limits_{n=1}^\infty a_nb_n$ by arithmetic of limts, and we're done.

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Oct 3rd 2011, 01:12 AM
  2. Proof of Convergence of a Series
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Jul 1st 2010, 06:22 AM
  3. Replies: 2
    Last Post: May 1st 2010, 09:22 PM
  4. Replies: 4
    Last Post: Dec 1st 2009, 03:23 PM
  5. A proof of convergence of an infinite series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 4th 2008, 08:23 AM

Search Tags


/mathhelpforum @mathhelpforum