Section 3: Geometric Series

A geometric sequence (or progression) is a sequence of the form $\{ar^{n-1}\}_{n=1}^\infty$. Here, $a$ is the first term and $r$ is the ratio. For example, $a_n = \dfrac{1}{2^{n-1}} = 1 \cdot \left(\dfrac{1}{2}\right)^{n-1}$. We can extract the ratio by dividing any term by the previous term: $r = \dfrac{ar^{n-1}}{ar^{n-2}}$.

Example: Find the 7th term of the geometric sequence: $-2, -18, -162, \ldots .$ Each term is being multiplied by $r = 9$, and the first term is $a = -2$, so $a_7 = -2 \cdot 9^6$.

Mini-Quiz: Find the 5th term of the sequence $3, 15, 75, \ldots$.

Let's talk about when geometric sequences converge. $$a_n = ar^{n-1} \begin{cases} \mbox{converges to } 0, & \mbox{if } |r| < 1, \mbox{ i.e., if } -1 < r < 1 \\ \mbox{converges to } a, & \mbox{if } r = 1 \\ \mbox{diverges to } \pm\infty, & \mbox{if } r > 1 \\ \mbox{diverges } & \mbox{if } r \leq -1. \end{cases}.$$

Now let's look at adding up geometric sequences and forming geometric series. We are considering $\sum_{n=1}^\infty ar^{n-1}$. If $|r| < 1$, then the series converges, and the sum is $$\sum_{n=1}^\infty ar^{n-1} = \frac{a}{1-r}.$$ This is the most important series to be familiar with.

Example: Let's test $\sum_{n=1}^\infty \frac{3}{2^{n-1}}$ and $\sum_{n=1}^\infty 2e^n$ for convergence. In the first series, the first term is $a = 3$, and the ratio is $r = 1/2$. So the sum is $a/(1-r) = 3/(1-1/2) = 3/(1/2) = 6$. In the second series, the first term is $a = 2e$, and the ratio is $r = e > 1$, so the series diverges.

Mini-Quiz: Test the series $\sum_{n=1}^\infty \dfrac{5}{3^{n-1}}$ and $\sum_{n=3}^\infty \dfrac{4}{(\ln{2})^n}$ for convergence. If they converge, what do they converge to?

A useful trick when the series doesn't start at 1 is to just think "first term is a, ratio is r."

Example: Find the sum $\sum_{n=4}^\infty \dfrac{2}{5^n}$. One way to approach this is to rewrite this as $\dfrac{1}{5^4}\sum_{n=1}^\infty \dfrac{2}{5^{n-1}}$. But it's easier to simply say the first term is $a = \dfrac{2}{5^4}$, and the ratio is $r = \dfrac{1}{5}$. So the sum is $a/(1-r) = \dfrac{2}{5^4(4/5)} = \dfrac{2}{5^4}\dfrac{5}{4} = \dfrac{1}{2\cdot 5^3}$.

Mini-Quiz: Find the sum $\sum_{n=7}^\infty \dfrac{2^n}{3^n}$.