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Homework Help: Limit of a geometric series

  1. Jun 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate the problem: [tex]\sum_{k=1}^\infty \frac{3^{(k-1)}}{4^{(k+1)}}[/tex]

    2. Relevant equations
    [itex]\displaystyle\lim_{n\rightarrow\infty} S_n = \frac{a}{1-r}[/itex]

    3. The attempt at a solution
    I know that the limit of the partial sequence is what i need to help solve this, but can't figure out how to do that. Can anyone point me in the right direction?
  2. jcsd
  3. Jun 20, 2012 #2
    Factor some terms out so it looks like (3/4)k, and k=0..∞

    EDIT: cleanest way might be pullout 1/16 and shift your index, an alternative might be pulling out 1/12 and adding (and subtracting) 1 (or 1/12 rather).

    SECOND EDIT: I apologize, I didn't read your question carefully.
    Last edited: Jun 20, 2012
  4. Jun 20, 2012 #3

    Ray Vickson

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    Start by figuring out what is r and what is a. Also, in [itex]S_n,[/itex] what values of n are included (i.e., where does n "start")?

  5. Jun 20, 2012 #4


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    Related to Ray Vickson's last point, the sum of a geometric series is
    [tex]\sum_{n=0}^\infty ar^n= \frac{a}{1- r}[/tex]

    Your sum is
    [tex]\sum_{n= 1}^\infty \frac{3^n}{4^n}= \sum_{n=1}^\infty \left(\frac{3}{4}\right)^n[/tex]

    Do you see the difference?
  6. Jun 20, 2012 #5
    Don't forget the 1/12 factor taken out in the HOI equation. :wink:
  7. Jun 20, 2012 #6
    Just as [itex](1-r)(1+r)=1-r^2[/itex], we have [itex](1-r)(1+r+r^2+\cdots+r^n)=1-r^{n+1}[/itex]. So if [itex]r\ne 1[/itex], then
    Additionally, if [itex]r<1[/itex], then we can take the limit of both sides. Notice the left side is your partial sum. Also, you may multiply both sides by [itex]a[/itex].
    Last edited: Jun 20, 2012
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