# B Index changes for series

1. Oct 2, 2016

### 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?

2. Oct 2, 2016

### mathman

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

3. Oct 2, 2016

### Mr Davis 97

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?

4. Oct 2, 2016

### Staff: Mentor

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

5. Oct 3, 2016

### Stephen Tashi

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.