1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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")?

    RGV
     
  5. Jun 20, 2012 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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
    [tex]1+r+r^2+\cdots+r^n=\frac{1-r^{n+1}}{1-r}.[/tex]
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Limit of a geometric series
  1. Geometric series (Replies: 1)

  2. Geometric Series (Replies: 4)

  3. Geometric series (Replies: 5)

  4. Series (Geometric?) (Replies: 5)

Loading...