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Series and Sequences

  1. Jul 5, 2008 #1
    Hope all of you had a good July 4th (for those that celebrate it)!

    Anyways, I'm trying to figure out series and sequences. I'm using Stewarts' Single Variable Calculus: Early Transcendentals, 6th. ed.

    For instance, under Section 11.4, the Comparison Tests, I don't understand how one arrives at bn.

    Example:

    [tex]\Sigma[/tex][tex]^{\infty}_{n=1}[/tex][tex]\frac{1}{2^{n}-1}[/tex]

    The book then proceeds to state:

    an = [tex]\frac{1}{2^{n}-1}[/tex], which I understand because it's written as [tex]\Sigma[/tex]an.

    However, the book then proceeds to state that:

    bn = [tex]\frac{1}{2^{n}}[/tex], which I have no idea how they got there. Do I just remove all constants from an? Or do I remove all variables that are not attached by an "n"?


    Also, my Calculus 1 is a bit rusty, but how exactly does (excerpted from p.717, Example 4 of Stewarts)

    [tex]\frac{(n+1)^{3}}{3^{n+1}}[/tex] [tex]\times[/tex] [tex]\frac{3^{n}}{n^{3}}[/tex]



    =[tex]\frac{1}{3}[/tex]([tex]\frac{n+1}{n}[/tex])3

    =[tex]\frac{1}{3}[/tex](1+[tex]\frac{1}{n}[/tex])3

    =[tex]\frac{1}{3}[/tex]
     
  2. jcsd
  3. Jul 5, 2008 #2
    i don't have a copy of that book, so here's a conjecture (going by the looks of it):
    that's the ratio test. you determine the ratio of (n+1)th term to the n th term as n approaches infinity, and conclude depending on whether the value obtained is greater or less that 1, the series is cgt or dvt. in this case it turns out to be 1/3.
     
  4. Jul 6, 2008 #3
    It's just implementing the comparison test:

    11.PNG
     
  5. Jul 6, 2008 #4
    Error: Does not compute.:confused:

    Sorry, I'm still confused :frown:
     
  6. Jul 7, 2008 #5
    I just skimmed your first post... I think this is relevant though:
    B sub n is just some other series that was (almost) arbitrarily picked because it converges.
     
  7. Aug 13, 2008 #6
    You can use the limit comparison test. I don't know a good mathematical way to explain this, but when I learned that test, my instructor called bn the "simpler version" of an. As it was stated earlier, we know that [tex]\sum^{\infty}_{n=0}[/tex][tex]\frac{1}{2^{n}}[/tex] converges. So, the limit comparison test says that if lim[tex]_{n\rightarrow\infty}[/tex][tex]\frac{a_{n}}{b_{n}}[/tex]=c, and c>0, then both an and bn converge or both diverge, if bn converges, so does an and the same goes for divergence.
     
  8. Aug 14, 2008 #7

    HallsofIvy

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    [tex]\frac{1}{2^n-1}> \frac{1}{2^n}[/tex]
    because the numerator on the left is smaller than the numerator on the right. However, that really doesn't help because you want to show this converges so you need "<" not ">" to a convergent series.

    You might say [tex]\frac{1}{2^n-1}< \frac{2}{2^n}[/tex]
    and then argue that since [tex]2\sum_{n=0}^\infty \frac{1}{2^n}[/tex] converges, so does [tex]\sum_{n=0}^\infty \frac{1}{2^n-1}[/tex]

    and, of course,
    [tex]\frac{(n+1)^3}{3^{n+1}}\frac{3^n}{n^3}= \frac{3^n}{3^{n+1}}\left(\frac{n+1}{n}\right)^3[/tex]
    [tex]= \frac{1}{3}\left(1+ \frac{1}{n}\right)^3[/tex]
     
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