Results 1 to 10 of 10

Math Help - convergence of a sequence

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    5

    convergence of a sequence

    Hi,

    I have a problem in my mind, it is as follows:

    We know that the sum of positive integer powers of a number x s.t. x is between 0 and 1 converges to 1/(1-x). Furthermore, assume that we have sequence b_{n} converging to b>0. Let

    s_{n} = {sum}_{i=0}^{n}[x^{n-i}*b_{i}] (in words: s_{n} equals to sum from i=0 to n of the terms x to the n-i multiplied by b_{i})

    The question is, does s_{n} converges?{may be helpful: if it converges, it converges to b/(1-x)}.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    675
    Thanks
    32
    We can show that s_{n+1}=xs_n+b_{n+1} by computing s_{n+1}-s_n.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2011
    Posts
    5
    Yes, i actually did that, then i think the following can be asserted:

    s_{n} converges if and only if s_{n+1} - s_{n} = (x-1)*s_{n} + b_{n+1} converges to 0. I run many MATLAB codes those implying the sequence indeed converges but i couldn't show it rigoruously.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    675
    Thanks
    32
    Let \displaystyle t_n:=b\sum_{j=0}^nx^j. If we denote by d_n:=s_n-t_n we get that d_{n+1}=xd_n+b_{n+1}-b hence |d_{n+1}|\leq|x||d_n|+|b_{n+1}-b|. Let l:=\limsup_{n}|d_n|; we have l\leq |x|l hence l=0.
    Last edited by girdav; March 24th 2011 at 01:58 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2011
    Posts
    5
    It's very nice actually, do you think we need to add that limsup(d_n)=liminf(d_n)=0? (since limit exists iff limsup=liminf)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by MADonmez View Post
    Hi,

    I have a problem in my mind, it is as follows:

    We know that the sum of positive integer powers of a number x s.t. x is between 0 and 1 converges to 1/(1-x). Furthermore, assume that we have sequence b_{n} converging to b>0. Let

    s_{n} = {sum}_{i=0}^{n}[x^{n-i}*b_{i}] (in words: s_{n} equals to sum from i=0 to n of the terms x to the n-i multiplied by b_{i})

    The question is, does s_{n} converges?{may be helpful: if it converges, it converges to b/(1-x)}.

    Thanks.
    Applying the Cauchy rule for the product of two converging series You have...

    \displaystyle \sum_{n=0}^{\infty} \sum_{i=0}^{n} x^{n-i}\ b_{i} = \sum_{n=0}^{\infty} x^{n}\ \sum_{i=0}^{\infty} b_{i} = \frac{b}{1-x}

    Kind regards

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

  7. #7
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    675
    Thanks
    32
    Quote Originally Posted by chisigma View Post
    Applying the Cauchy rule for the product of two converging series You have...

    \displaystyle \sum_{n=0}^{\infty} \sum_{i=0}^{n} x^{n-i}\ b_{i} = \sum_{n=0}^{\infty} x^{n}\ \sum_{i=0}^{\infty} b_{i} = \frac{b}{1-x}

    Kind regards

    \chi \sigma
    We know that the sequence \left\{b_n\right\} is convergent (to b) but not the series.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    675
    Thanks
    32
    Quote Originally Posted by MADonmez View Post
    It's very nice actually, do you think we need to add that limsup(d_n)=liminf(d_n)=0? (since limit exists iff limsup=liminf)
    Yes it's a step of our proof, even if someone who read until " l=0" understands that the limit exists. We have to add that \left\{t_n\right\} is convergent and \left\{s_n\right\} is also convergent and has the same limit.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Mar 2011
    Posts
    5
    thanks, but b_{n} is a "sequence" converging to b, so the series above diverges
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Mar 2011
    Posts
    5
    thank you very much
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Convergence of a sequence
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: August 30th 2010, 11:51 AM
  2. Replies: 7
    Last Post: October 12th 2009, 10:10 AM
  3. Sequence convergence
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: October 6th 2009, 07:46 PM
  4. Replies: 6
    Last Post: October 1st 2009, 09:10 AM
  5. Replies: 6
    Last Post: October 24th 2008, 01:45 PM

Search Tags


/mathhelpforum @mathhelpforum