Partial Sum of Geometric Series

Theorem

The sum of the first \(n\) terms of a geometric series given by initial term \(a\) and common ratio \(r\) is

\[ \frac{a(1 - r^n)}{1 - r}.\]

Proof

Consider the sum

\[ \sum_{k = 0}^{n - 1} a r^k = a + ar + \dots + ar^{n - 1}.\]

Finding a general expression for the limit of such a sum is just an exercise in re-indexing sums. We proceed as follows

\[\begin{align*} S &= \sum_{k = 0}^{n - 1} a r^k \\ &= a \sum_{k = 0}^{n - 1} r^k \\ &= a + a \sum_{k = 1}^{n - 1} r^k \\ &= a + ar \sum_{k = 1}^{n - 1} r^{k - 1} \\ &= a + ar \sum_{k = 0}^{n - 2} r^k \\ &= a + ar \left(\left(\sum_{k = 0}^{n - 2} r^k\right) + r^{n - 1} - r^{n - 1} \right) \\ &= a + ar \left(\sum_{k = 0}^{n - 1} r^k\right) - ar^n \\ &= a + r \left(\sum_{k = 0}^{n - 1} ar^k\right) - ar^n \\ &= a + r S - ar^n \\ (1 - r)S &= a - ar^n \\ S &= \frac{a(1 - r^n)}{1 - r}. \\ \end{align*}\]