Does Changing Indices Affect the Formula for a Geometric Series?

[Mr Davis 97]

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?

 

[mathman]

$\sum_{k=0}^{n-1}(\frac{1}{2})^{k+1}=\frac{1}{2}\sum_{k=0}^{n-1}(\frac{1}{2})^{k}$

 

[Mr Davis 97]

$\sum_{k=0}^{n-1}(\frac{1}{2})^{k+1}=\frac{1}{2}\sum_{k=0}^{n-1}(\frac{1}{2})^{k}$

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?

 

[Mark44]

$\sum_{k=0}^{n-1}(\frac{1}{2})^{k+1}=\frac{1}{2}\sum_{k=0}^{n-1}(\frac{1}{2})^{k}$

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?

Changing the indices doesn't change the value that the summation converges to, assuming the sum converges.

 

[Stephen Tashi]

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?

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.