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Determine the sum of the given series:

  1. Jul 22, 2014 #1
    1. The problem statement, all variables and given/known data

    Sum starting from n=1 to infinity for the expression, (3/4^(n-2))

    What the solutions manual has done is multiply the numerator and the denominator by 4.

    12/(4^(n-1))

    I don't know what they have done from here on:

    12 / (1 - 1/4)

    = 16

    Can someone explain the solution? Thank you.
     
  2. jcsd
  3. Jul 22, 2014 #2
    It's a geometric series.
     
  4. Jul 22, 2014 #3
    I'm having trouble seeing how to use the fact that it is a geometric series in the problem. Could you explain further?
     
  5. Jul 22, 2014 #4
    Do you have a formula for the sum of a geometric series?
     
  6. Jul 22, 2014 #5
    I know for a geometric series the formula is Sn = (a(1-r^n)) / 1-r and for an infinite geometric series S = a/1-r
     
  7. Jul 22, 2014 #6
    OK then ... does ##\frac{12}{1-\frac{1}{4}}## look like one of those forms?
     
  8. Jul 22, 2014 #7
    Well the rate is 4, and the first value, a, is 1/4. Also 12 fits no where into any of these formulas so 12/(1-1/4) doesn't look like either of the two formulas for geometric series.
     
  9. Jul 22, 2014 #8
    Oh wait they are treating the entire expression as a geometric series not just the denominator. Is that it? Give me a second to see then.
     
  10. Jul 22, 2014 #9
    Ok I get it now I am finally getting 16 as an answer. My question is now why don't they divide 16 by 4 to get the original value to which the series converges to. (because in the beginning they multiplied the entire expression by 4?)
     
  11. Jul 22, 2014 #10
    They multiplied both the numerator and denominator of the term inside the sum by 4, which is the same as multiplying the whole term inside the sum by 4/4=1. Multiplying by 1 doesn't change anything, so there is nothing to "undo". It goes $$\sum\limits_{n=1}^\infty\frac{3}{4^{n-2}}=\sum\limits_{n=1}^\infty1\cdot\frac{3}{4^{n-2}}=\sum\limits_{n=1}^\infty\frac{4}{4}\cdot\frac{3}{4^{n-2}}=\sum\limits_{n=1}^\infty\frac{12}{4^{n-1}}=\sum\limits_{n=1}^\infty12\cdot\frac{1}{4^{n-1}}=\sum\limits_{n=1}^\infty12\cdot(\frac{1}{4})^{n-1}$$ with the formula for the geometric series applied to the RHS.
    Everything's equal, so ...
     
  12. Jul 22, 2014 #11
    Oh yes that is true the constant that is being multiplied is 1 not 4, so it won't change anything. Thank you for your help throughout this process; it is greatly appreciated!
     
  13. Jul 22, 2014 #12

    HallsofIvy

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    [tex]\sum_1^\infty \frac{3}{4^{n-2}}= \sum_1^{\infty} \frac{3}{16}\frac{1}{4^n}= \frac{3}{16}\sum_1^\infty \frac{1}{4^n}[/tex]

    BUT the formula for the sum of a geometric series: [tex]\sum_{n=0}^\infty a^n= \frac{1}{1- a}[/tex] starts with n= 0, not n= 1. There are two ways to handle that.
    1) factor a "1/4" out of the series:
    [tex]\frac{3}{64}\sum_{n=1}^\infty \frac{1}{4^{n-1}}= \frac{3}{64}\sum_{i= 0}^\infty \frac{1}{4^i}[/tex] where i= n-1.

    2) Do the entire geometric series from 0 to infinity, then subtract off the n= 0 term:
    [tex]\frac{3}{4}\sum_{n=0}^\infty \frac{1}{4^n}- \frac{3}{4}[/tex]
    since when n= 0, [tex]\frac{1}{4^n}= 1[/tex].
     
    Last edited: Jul 22, 2014
  14. Jul 22, 2014 #13
    Some texts give ##\sum\limits_{n=1}^\infty ar^{n-1}=\frac{a}{1-r}## as (one of) the formula(s) for the geometric series. Given the answer from the solutions manual, I'm guessing the topic creator is using one of those texts.
     
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