- #1
Mr Davis 97
- 1,462
- 44
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?
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?