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Geometric Progression

  1. Apr 8, 2009 #1
    what is the nth partial sum of


    i don't understand why i can't do this?

    i have [itex]\sum_{k=1}^{n} ar^k=\frac{a(r^{n+1}-r)}{r-1}[/itex]

    ok but then when i sub in i get messed up i get [itex]\frac{(-x)^{n+1}+x}{-(1+x)}[/itex] is there any way of simplifying this? also when i set n=1 i get that the sum is -x when it should be 1-x???
  2. jcsd
  3. Apr 8, 2009 #2
    Do you know how to derive the nth partial sum of 1 + x + x^2 + ... ?
  4. Apr 8, 2009 #3
    off the top of my head isn't it something like

    [itex](1-x)(1+x+x^2+...+x^n)=1-x^{n+1} \Rightarrow 1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}[/itex]
  5. Apr 8, 2009 #4
    also, another thing i'm having trouble seeing is why [itex]\sum \frac{1}{n+1}[/itex] diverges.
    my initial instinct was to use the copmarison test with [itex]\sum \frac{1}{n}[/itex] as this diverges.

    so if i can show [itex]\frac{1}{1+n} \geq \frac{1}{n}[/itex] then all is well but in fact [itex]\frac{1}{n} \geq \frac{1}{1+n}[/itex] e.g. set n=1 to see

    what's going on here?
  6. Apr 8, 2009 #5
    How can you modify this so that it works for [itex]1 - x + x^2 - \cdots[/itex]

    There's no need for the Comparison Test. Just note that

    [tex]\sum_{n=1}^\infty \frac{1}{n + 1} = \frac{1}{2} + \frac{1}{3} + \cdots = \sum_{n=2}^\infty \frac{1}{n}[/tex]
  7. Apr 9, 2009 #6
    would it have a numerator of [itex]1-(-1)^nx^{n+1}[/itex]?
  8. Apr 9, 2009 #7
    I don't know this of the top of my head, but if you write up what you did to get that, I can look it over.
  9. Apr 9, 2009 #8


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    WHAT would? You started talking about
    [tex]\sum_{i= 0}^n (-x)^i[/itex]
    and then changed to
    [tex]\sum_{n=0}^\infty \frac{1}{n+1}[/tex]

    Are you back to the first question again?
  10. Apr 10, 2009 #9
    Looking at your equation right here, I notice one problem that would fix it rather quickly. When you tried to find the nth partial sum, you are took:

    [itex]\sum_{k=1}^{n} (-x)^k[/itex]

    The only problem with this is that when n = 1, you get -x just like you got. You accidentally dropped the first term when you switched to sum notation. Just change it to:

    [itex]\sum_{k=0}^{n} (-x)^k[/itex]

    This way, 1 is the first term. Your new equation is:

    [itex]\sum_{k=0}^{n} a*r^k = \frac{a(r^{n + 1} - 1)}{r - 1}[/itex]
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