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B Index changes for series

  1. Oct 2, 2016 #1
    I have always been a bit confused about how changing indices in a summations changes the resulting closed formula.

    Take this geometric series as an example: ##\displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^n}##. Putting it into summation notation, we have ##\displaystyle \sum_{k = 1}^{n} (\frac{1}{2})^k##. Converting a little bit, we get ##\displaystyle \sum_{k = 1}^{n} \frac{1}{2} (\frac{1}{2})^{k - 1}##, which fits nicely into the formula for a geometric series: ##\displaystyle \frac{\frac{1}{2} (1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}}##.

    However, what if we wanted to change the indices so that we start from zero? Then we would have ##\displaystyle \sum_{k = 0}^{n - 1} (\frac{1}{2})^{k + 1}##. How would we convert this expression to get a closed form for the series?
     
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  3. Oct 2, 2016 #2

    mathman

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    Gold Member

    [itex]\sum_{k=0}^{n-1}(\frac{1}{2})^{k+1}=\frac{1}{2}\sum_{k=0}^{n-1}(\frac{1}{2})^{k}[/itex]
     
  4. Oct 2, 2016 #3
    Would this then be ##\displaystyle \frac{\frac{1}{2} (1 - (\frac{1}{2})^{n}) }{1 - \frac{1}{2}}##? Is always the case that changing the indices results in the same formula?
     
  5. Oct 2, 2016 #4

    Mark44

    Staff: Mentor

    Changing the indices doesn't change the value that the summation converges to, assuming the sum converges.
     
  6. Oct 3, 2016 #5

    Stephen Tashi

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    Changing the indices won't change the formula for the sum provided you don't change the meaning of the variables in the formula for the sum. In your example , "n" keeps the same meaning when you change indices.
     
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