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Infinite series without k-1

  1. Apr 28, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm told to evaluate the following to the thousandths place:

    [tex]\Sigma[/tex] 7*(0.35)^k

    2. Relevant equations
    We know that an infinite equation can be expressed as:


    3. The attempt at a solution

    The first term (a1) is 7 and r=.35 so I can plug those into the above equation. I can see that we're doing something along the lines of:


    However, there are two issues I'm having.

    1) How do I plug in n when it's infinity?

    2) In the non-infinite sum of geometric series problems I've worked where in the [tex]\Sigma[/tex] equation it's r^k versus the normal format of r^k-1 I had to modify the problem to make it k-1 and ended up multiplying the whole thing by r and then subtracting r from the whole thing... It was kind of crazy and I'm still not sure why and what happened.

    Do I have to do something like that here?

    How do I solve this equation?

    Thank you for your time!
  2. jcsd
  3. Apr 28, 2010 #2


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    Science Advisor

    No, this is wrong. The formula for a finite geometric sum is
    [tex]\sum_{k=0}^n ar^k= \frac{a(1- r^{n+ 1})}{1- r}[/tex].

    You get the formula for the infinite series by taking the limit as n goes to infinity. It only converges if -1< r< 1 and in that case [itex]r^n[/itex] goes to 0. The sum of an infinite geometric series,
    [tex]\sum_{n=0}^\infty ar^n= \frac{a}{1- r}[/tex]

    There is no "n" in the formula for the sum of an infinite series.

    No, you do not.

    Use the formula
    [tex]\sum_{n=0}^\infty ar^n= \frac{a}{1- r}[/tex]
  4. Apr 28, 2010 #3
    Alright, thanks!

    Now I guess my issue is that k=1 in my equation and I'm really not sure how to get it to 0.
  5. Apr 28, 2010 #4
    Just for future reference if anyone has this question again.

    I figured it out after HallsofIvy's help, thanks!

    You just take the result of a/(1-r) which happens to be 10.769 in this case and subtract 1*a or '7' to get the end result of 3.769.
  6. Apr 29, 2010 #5


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    Science Advisor

    The "k= 0" term in [itex]7(0.35)^k[/itex] is [itex]7(0.35)^0= 7[/itex]. To get the sum from k= 1 to infinity. ust use the formula from 0 to infinity, the subtract 7.

    Or: let n= k-1 so that when k= 1, n= 0. Then k= n+ 1 so [itex]7(0.35)^k= 7(0.35)^{n+1}= 7(0.35)(0.35)^n[/math]. Use the formula from n=0 to infinity with [math]a_0= 7(0.35)= 2.45 and r still 0.35.

    Both of those should give the same answer.
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