Results 1 to 3 of 3

Math Help - sequences

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    19

    sequences

    Hi all,
    I'm trying to solve the following 2 questions , can anyone help?
    1. Let for k . Show that there exists a sequence 0 such that

    2.Let .Show that there exists a sequence such that .

    For the first part , I think for is suitable. Can anyone give a hint for the second part?
    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
    If \displaystyle \sum_{k=1}^{\infty} |a_{k}|< \infty, then the series converges absolutely and that means that a_{k} \rightarrow 0 and for k 'large enough' is \displaystyle |a_{k}|< \frac{1}{k^{\alpha}} where \alpha>1. Now if we suppose that \forall k, a_{k}\ne 0 and set b_{k}= \ln |a_{k}| it will be |b_{k}| \rightarrow \infty and for k 'large enough' \displaystyle |a_{k}\ b_{k}|< \frac{1}{k^{\beta}} where \beta>1 so that \displaystyle \sum_{k=1}^{\infty} |a_{k}\ b_{k}|< \infty ...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by chisigma View Post
    If \displaystyle \sum_{k=1}^{\infty} |a_{k}|< \infty, then the series converges absolutely and that means that a_{k} \rightarrow 0 and for k 'large enough' is \displaystyle |a_{k}|< \frac{1}{k^{\alpha}} where \alpha>1.
    Unfortunately that is not true. There exist convergent series \sum a_n of positive numbers, such that the condition a_n<\frac1{n^\alpha} does not hold for all 'large enough' n, for any \alpha>1. For example, if a_n = \dfrac1{n(\ln n)^2}, then \sum a_n converges (by the integral test), but n^\alpha a_n\to\infty for all \alpha>1.

    Quote Originally Posted by hermanni View Post
    Let .Show that there exists a sequence such that .
    There is a standard trick to construct this example. Let t_n denote the tail of the series \sum|a_k|, in other words t_n = \sum_{k=n}^\infty|a_k|. Then t_n\to0 as n\to\infty. Let b_k = 1/\sqrt{t_k}. Clearly b_k\to\infty as k\to\infty.

    For the clever proof that \sum|a_kb_k| converges, see here.
    Last edited by Opalg; February 14th 2011 at 05:37 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Convergence in sequences of sequences
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: October 19th 2010, 08:28 AM
  2. Sequences and the sequences' arithmetics
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 6th 2010, 10:31 PM
  3. sequences
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: January 30th 2010, 04:17 PM
  4. Monotone sequences and Cauchy sequences
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: March 21st 2009, 09:59 PM
  5. Replies: 5
    Last Post: January 16th 2008, 05:51 PM

Search Tags


/mathhelpforum @mathhelpforum