Infinite Geometric Series

Bouncing Ball

Suppose a tennis ball is dropped from a height of $1 \text{ meter}$ . In an idealized model, each bounce reaches $\frac{1}{4}$ of the previous height.

The release height followed by the rebound heights forms a geometric sequence:

1, \quad 1 \times \frac{1}{4}, \quad (1 \times \frac{1}{4}) \times \frac{1}{4}, \quad \dots

The first term ( $a$ ) is $1$ , and the ratio ( $r$ ) is $\frac{1}{4}$ .

What total distance does the ball travel from the moment it is released until it comes to rest?

In the idealized model, the ball moves down, then up, down again, and so on without a final bounce. The total distance is built from all downward and upward paths. Because their lengths form an infinite geometric pattern, the calculation uses an infinite geometric series.

The Ratio Condition for Convergence

The mathematical series does not have a final bounce. Instead, its partial sums may approach a finite limit. For a geometric series, this happens when the absolute value of the ratio $r$ is less than $1$ .

|r| < 1 \quad \text{or} \quad -1 < r < 1

A series like this is called convergent, meaning its partial sums approach a specific finite value. In the ball example, $r = \frac{1}{4}$ , and since $-1 < \frac{1}{4} < 1$ , the series is convergent.

If $|r| \ge 1$ and $a\ne0$ , the terms do not approach zero, so the series is divergent. Its partial sums may grow without bound, decrease without bound, or oscillate. Divergence means that no finite limit exists. It does not always mean $+\infty$ or $-\infty$ .

Calculating the Sum of an Infinite Series

How do we calculate the sum of a convergent infinite series? We start with the formula for the sum of the first $n$ terms of a geometric series:

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

For an infinite series, we examine $S_n$ as $n$ grows without bound. If the series is convergent ( $-1 < r < 1$ ), then $r^n$ approaches zero as $n$ approaches infinity. For example, $(\frac{1}{4})^2 = \frac{1}{16}$ , $(\frac{1}{4})^3 = \frac{1}{64}$ , and $(\frac{1}{4})^{10} \approx 0.00000095$ . The larger $n$ becomes, the closer $(\frac{1}{4})^n$ gets to $0$ .

So, for $n \to \infty$ and $-1 < r < 1$ , we have $r^n \to 0$ . The formula becomes:

\begin{aligned} S_\infty &= \lim_{n \to \infty} S_n \\ &= \lim_{n \to \infty} \frac{a(1 - r^n)}{1 - r} \\ &= \frac{a(1 - 0)}{1 - r} \end{aligned}

The formula for the sum of a convergent infinite geometric series is:

S_\infty = \frac{a}{1 - r}

In this notation:

$S_\infty$ = sum of the infinite series
$a$ = first term
$r$ = ratio, with $-1<r<1$

A Check with Halved Distances

Consider the series:

1+\frac12+\frac14+\frac18+\cdots

Here $a=1$ and $r=\frac12$ , so:

S_\infty=\frac{1}{1-1/2}=2

No finite partial sum equals $2$ , but the partial sums can get as close to $2$ as desired. The infinite sum is the limit of those finite partial sums.

Calculating the Total Distance

We can calculate the total distance traveled by the ball using the $S_\infty$ formula. There are two parts to the path:

Downward Path: The ball falls from height $a$ , then falls again after the first bounce ( $ar$ ), falls again after the second bounce ( $ar^2$ ), and so on.
- Downward series: $a + ar + ar^2 + ar^3 + \dots$
- First term ( $a_{\downarrow}$ ) is $a$
- Ratio ( $r_{\downarrow}$ ) is $r$
- Sum of downward path: $S_{\downarrow}=\frac{a_{\downarrow}}{1-r_{\downarrow}}=\frac{a}{1-r}$
Upward Path: The ball moves up after the first bounce ( $ar$ ), up again after the second bounce ( $ar^2$ ), and so on.
- Upward series: $ar + ar^2 + ar^3 + \dots$
- First term ( $a_{\uparrow}$ ) is $ar$
- Ratio ( $r_{\uparrow}$ ) is $r$
- Sum of upward path: $S_{\uparrow}=\frac{a_{\uparrow}}{1-r_{\uparrow}}=\frac{ar}{1-r}$

The total distance is the sum of all downward distances and all upward distances.

\begin{aligned} \text{Total} &= S_{\downarrow} + S_{\uparrow} \\ &= \frac{a}{1 - r} + \frac{ar}{1 - r} \end{aligned}

For the tennis ball example with $a = 1 \text{ meter}$ and $r = \frac{1}{4}$ :

\begin{aligned}\text{Total} &= \frac{1(1 + \frac{1}{4})}{1 - \frac{1}{4}} \\ &= \frac{1(\frac{5}{4})}{\frac{3}{4}} \\ &= \frac{5/4}{3/4} \\ &= \frac{5}{4} \times \frac{4}{3} \\ &= \frac{5}{3}\end{aligned}

So, the model gives a total travel distance of $\frac{5}{3} \text{ meters}$ .

This is an idealized mathematical model. A real ball does not complete infinitely many measurable bounces, and its bounce ratio is not perfectly constant.

A Second Calculation: The total distance can also be calculated as the first downward path plus $2 \text{ times}$ the sum of all upward paths.

\begin{aligned} \text{Total} &= a + 2 \times S_{\uparrow} \\ &= a + 2 \times \left( \frac{ar}{1 - r} \right) \end{aligned}

Both methods give the same result.

Practice with an Alternating Series

Find the sum of:

12-6+3-\frac32+\cdots

Here $a=12$ and $r=-\frac12$ . Since $|r|<1$ , the series converges:

S_\infty=\frac{12}{1-(-1/2)}=8

The negative ratio makes the terms alternate in sign, but their magnitudes shrink quickly enough for the partial sums to approach $8$ .