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Summation ?

  1. Aug 20, 2008 #1
    1. The problem statement, all variables and given/known data
    I am to find the sum of the series, but what do i do if it is infinite?? no clue.
    i'm also not sure how to type the symbols so i hope you can understand me:shy:
    : (Sum) n=0, limit = infinity: 1/2(2/3)^n

    2. Relevant equations
    i 'm not sure.


    3. The attempt at a solution
    i understand how to add up to a limit, but what do i do with the infinity?
     
  2. jcsd
  3. Aug 20, 2008 #2
    Well if the 1/2 is the first term in the product then you can pull it out front (this is obvious by the distributive law). Then recognize it as a geometric series. There is a simple formula for an infinite geometric series which you should try deriving yourself.
     
  4. Aug 20, 2008 #3
    .. 1/2(sum)n=0, infinity (2/3)^n+1 ..
    ?? do you just leave it like that?..
    as a sort of equation?
     
  5. Aug 20, 2008 #4
    Ok an infinite geometric series like a normal one has a first term conventionally detonated as [tex]a[/tex] and common ratio [tex]r[/tex]. Since we start at n = 0, the first term is (2/3)^0 = 1 (You can't just change the exponent from n to n+1 unless you change the starting value. Making do with what you're given is best in this case). Then we go to n = 1, the second term is (2/3)^1 = 1*(2/3) = 2/3. Then for n = 2, the third term is (2/3)^2 = (2/3)(2/3). Now the idea is to sum all of these terms, i.e. 1 + 1*(2/3) + 1*(2/3)*(2/3) + 1*(2/3)(2/3)(2/3) + ...

    So we can generalize a bit. We have [tex]a[/tex] as our first term and a common ratio [tex]r[/tex]. Our sum, which we'll denote [tex]S[/tex] is

    [tex]S = a + ar + ar^2 + ar^3 + ...[/tex] (1)

    (Compare with the first paragraph to understand why this is true).

    Now we need to solve for [tex]S[/tex] because that gives us the sum. The trick here is to multiply [tex]S[/tex] by our common ratio [tex]r[/tex], i.e. multiply both sides of (1) above by [tex]r[/tex] and write it underneath. Then subtract the new equation from (1) and see for yourself how all the terms cancel and allow you to solve easily for [tex]S[/tex]
     
  6. Aug 20, 2008 #5

    tiny-tim

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    Hi Wholewheat458! :smile:

    [tex]\sum_{n=0}^{\infty} \frac{1}{2}\,\left(\frac{2}{3}\right)^n[/tex]

    [tex]=\ \frac{1}{2}\,\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n[/tex]

    You should know what [tex]\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n[/tex] is,

    but if you don't, just sum it from 0 to N, and then let N --> ∞ :smile:

    (btw, it's [noparse][tex]\sum_{n=0}^{\infty} \frac{1}{2}\,\left(\frac{2}{3}\right)^n[/tex][/noparse] :wink:)
     
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