Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Looking for a Sequence

  1. Jun 28, 2011 #1
    I'm looking for a convergent sequence [tex]s_n[/tex] such that:


    I've already gone pretty far afield in my hunt for such a sequence, so I thought I'd enlist the help of you fine folks in my search.
  2. jcsd
  3. Jun 28, 2011 #2


    User Avatar
    Science Advisor

    If you are willing to accept conditionally convergent series, try sn=(-1)n/√n.
  4. Jun 28, 2011 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    He is talking about a sequence, not a series. At least that's what he said although he did use notation that suggests partial sums of a series. Your sequence converges (not conditionally). If he meant what he said, your suggestion may work depending on what he really wants. If n is even, n(sn-sn-1) >0 and goes to ∞ through the subsequence of even numbers.But if n is odd it → -∞. Perhaps that is good enough.
  5. Jun 28, 2011 #4
    I think mathman may have misspoke when he said series, his sequence will work, all I needed was the absolute value of n(s_n-s_n-1) to increase without bound. Although for my ultimate purpose it turns out to be insufficient.

    I'm taking a look at Cesaro summation: [tex]\sigma_n=\frac{s_0+s_1+...+s_n}{n+1}[/tex]
    and trying to find a case where [tex]\lim\sigma_n=\sigma[/tex] while [tex]\lim s_n\neq\sigma.[/tex]
    And if I am interpreting my book correctly the condition for the sequence s_n given in my original post is a necessary condition for this to occur, but upon doing the computation with mathman's example it appears not sufficient:
    [tex]Set\;\; s_0=0,\;\; define\;\; s_n=\sum_{k=1}^{n}\frac{(-1)^k}{\sqrt{k}}\;\; for\;\; n\geq 1,\;\; and\;\; assume\;\; that\;\; \sigma_n=\sigma.\;\;\;\; Then\;\; [/tex][tex]\sigma_n=\frac{1}{n+1}\sum_{i=1}^{n}\frac{(-1)^i(n-i+1)}{\sqrt{i}}[/tex][tex]=\frac{1}{n+1}\sum_{i=1}^{n}\frac{(-1)^i(n+1)}{\sqrt{i}}-\frac{1}{n+1}\sum_{i=1}^{n}\frac{(-1)^ii}{\sqrt{i}}[/tex][tex]=\sum_{i=1}^{n}\frac{(-1)^i}{\sqrt{i}}-\frac{1}{n+1}\sum_{i=1}^{n}\frac{(-1)^ii}{\sqrt{i}}[/tex][tex]=s_n+\frac{1}{n+1}\sum_{i=1}^{n}(-1)^i\sqrt{i}\rightarrow \lim s_n=\sigma\;\; as\;\; n\rightarrow\infty.[/tex]
    Where the penultimate equality follows by definition, and the ultimate by our assumption.

    So I'm curious if either of you can come up with a sequence which does provide different limits for sigma_n and s_n.

    P.S. does anyone know how to not automatically go down to the next line every time one codes in some latex, it used to not do that but I think physicsforum like changed their latex format and now it does, thanks.

    Edit: Actually I'm not entirely sure that [tex]\lim_{n\rightarrow\infty}\frac{1}{n+1}\sum_{i=1}^{n}(-1)^i\sqrt{i}=0.[/tex]
    If not than mathman's sequence does work after all.
    Last edited: Jun 29, 2011
  6. Jun 29, 2011 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I assume you are aware that Cesaro summation is convergence preserving. Look at the sequence {an} where an = 1 if n is even and 0 if n is odd. Here σn → 1/2 while {an} diverges.
  7. Jun 29, 2011 #6
    By convergence preserving are you implying that if both sigma_n and s_n converge than they must converge to the same value?
  8. Jun 29, 2011 #7


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes. If a sequence converges then its Cesaro sum converges to the same limit. There is a general theorem by Toeplitz giving a conditions on a summability matrix that make it convergence preserving, which the Cesaro sum matrix satisfies. You can read about it at


    It is also not difficult to prove that result for Cesaro summation directly.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Looking for a Sequence
  1. Proving sequences (Replies: 12)

  2. Cauchy Sequences (Replies: 4)

  3. Puzzled by sequence (Replies: 9)