Introduction

Welcome to this guide on geometric series! This article will guide you through the following:

Definition

Let us first recap the basic definition of a geometric sequence:

Consider this sequence $3,6,12,24,\mathrm{48...}$

It is easy to see that there is a common factor of $2$ between every term. Thus we can rewrite the sequence like this:

$3\cdot 2^0,3\cdot 2^1,3\cdot 2^2,3\cdot 2^3,3\cdot 2^4$

In general, the $n$th term expression can be written as (where the first term is $a_0$):

$a_n=3\cdot 2^n$

If we take the sum of all terms of the geometric sequence, we arrive at what we call a geometric series. Geometric series have a property where the quotient of two neighboring terms of a geometric series is a constant, which is known as the common ratio ($2$ in the previous example).

With this knowledge, we can generalize the $n$th term expression given before:

$a_n=a_0\cdot r^n$

Where $a_n$ is the $n$th term, $a_0$ is the first term, $r$ is the common ratio.

It should be noted that AP Calculus BC uses zero-based indexing when discussing geometric series in their Course Exam Description, which we shall thus primarily be using. Again, please take care to note what the starting index is.  

Convergence of Geometric Series

Consider a geometric sequence of $\sum _{n=0}^{\infty }a\cdot r^n$

Geometric series diverge if $|r|\ge 1$, and converge if $0<|r|<1$.

If the series converges, it converges to a value of: 

$S=\sum _{n=0}^{\infty }a{r}^n=\frac{a}{1-r}$

Again, please note that zero-based indexing is used for the above formula. The formula, when represented in one-based indexing, is as follows:

$S=\sum _{n=1}^{\infty }{a}_1{r}^{n-1}=\frac{a_1}{1-r}$

Proof

You don't need to know this for the AP exam, but this section is here for the completeness of knowledge. Let $S$ be the number the geometric series converge to:

$S_n=a+ar+a{r}^2+...a{r}^{n-1}$

$r{S}_n=r(a+ar+a{r}^2+...a{r}^{n-1})$

$r{S}_n=ar+a{r}^2+a{r}^3+...+a{r}^{n-1}+a{r}^n$

$S_n-r{S}_n=a-a{r}^n$

$S_n(1-r)=a(1-r^n)$

$S_n=\frac{a(1-r^n)}{1-r}$

Since this is an infinite series we are dealing with, we'll let $n\to \infty$. Considering that we specifically like geometric series where $0<|r|<1$, we then see $\underset{n\to \infty }{\lim }{r}^n=0$

Thus, we realize that $\underset{n\to \infty }{\lim }(1-r^n)=1$

Substituting it back into the original expression, we realize then:

$\therefore S_n=\frac{a}{1-r}$

Worked Examples

Example 1:

Determine if this series converges, $\sum _{n=0}^{\infty }1.1^n$.

Solution: Since the common ratio $r=1.1>1$, this series diverges.

Example 2:

$\sum _{n=0}^{\infty }3{\left(\frac{1}{2}\right)}^n$

Solution: since $\frac{1}{2}<1$, this series converges, the value it converges to is

$\sum _{n=0}^{\infty }3{\left(\frac{1}{2}\right)}^n=\frac{3}{1-\frac{1}{2}}=6$

Practice

Determine if the following series converge, if it converge, find the value it converges to