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URL: https://nakafa.com/en/subjects/mathematics/sequence-series/infinite-geometric-series
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Learn infinite geometric series with bouncing ball examples, then study convergence, divergence, and infinite sum formulas.

---

## Bouncing Ball

Imagine throwing a tennis ball from a height of $$1 \text{ meter}$$. The ball will bounce, but each bounce height is only $$\frac{1}{4}$$ of the previous bounce height.

Visible text: Imagine throwing a tennis ball from a height of . The ball will bounce, but each bounce height is only of the previous bounce height.

The bounce heights form a **geometric sequence**:

Component: MathContainer
Children:

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

```math
1, \quad \frac{1}{4}, \quad \frac{1}{16}, \quad \frac{1}{64}, \quad \dots
```

With the first term ($$a$$) being $$1$$ and the ratio ($$r$$) being $$\frac{1}{4}$$.

Visible text: With the first term () being and the ratio () being .

Now, think about this: what is the _total distance_ traveled by the ball from the moment it's thrown until it finally stops?

The ball moves down, then up, down again, up again, and so on, until it stops. The total distance is the sum of all downward paths and all upward paths. Since the ball keeps bouncing (although lower and lower), we are summing an infinite number of distances. This is called an **Infinite Geometric Series**.

### When Does the Series Stop?

Logically, the ball will eventually stop bouncing, right? This happens because the bounce height gets smaller and smaller, approaching zero. In mathematics, this occurs if the absolute value of the ratio ($$r$$) is less than $$1$$.

Visible text: Logically, the ball will eventually stop bouncing, right? This happens because the bounce height gets smaller and smaller, approaching zero. In mathematics, this occurs if the absolute value of the ratio () is less than .

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

A series like this is called **convergent**, meaning its sum approaches a specific
finite value (not infinity). In the ball example, $$r = \frac{1}{4}$$
, and since $$-1 < \frac{1}{4} < 1$$, the series is
convergent. The ball will stop, and its total distance can be calculated.

Visible text: A series like this is called **convergent**, meaning its sum approaches a specific
finite value (not infinity). In the ball example, 
, and since , the series is
convergent. The ball will stop, and its total distance can be calculated.

If $$|r| \ge 1$$ (i.e., $$r \ge 1$$ or $$r \le -1$$), the bounce height won't decrease or might even increase. The series is called **divergent**, and its sum is infinite ($$\pm \infty$$).

Visible text: If (i.e., or ), the bounce height won't decrease or might even increase. The series is called **divergent**, and its sum is infinite ().

## 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:

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

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

For an infinite series, we look for the value of $$S_n$$ as $$n$$ becomes
very large (approaches infinity). If the series is convergent ($$-1 < r < 1$$
), then the value of $$r^n$$ will approach 0 as $$n$$ approaches
infinity. Example: $$(\frac{1}{4})^2 = \frac{1}{16}$$, $$(\frac{1}{4})^3 = \frac{1}{64}$$
, $$(\frac{1}{4})^{10} \approx 0.00000095$$. The larger $$n$$ gets,
the closer $$(\frac{1}{4})^n$$ gets to $$0$$.

Visible text: For an infinite series, we look for the value of as becomes
very large (approaches infinity). If the series is convergent (
), then the value of will approach 0 as approaches
infinity. Example: , 
, . The larger gets,
the closer gets to .

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

Visible text: So, for and , we have . The formula becomes:

```math
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}
```

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

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

Where: * $$S_\infty$$ = Sum of the infinite series * $$a$$ =
First term * $$r$$ = Ratio ($$-1 < r < 1$$)

Visible text: Where: * = Sum of the infinite series * =
First term * = Ratio ()

## 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:

Visible text: We can calculate the total distance traveled by the ball using the formula. There are two parts to the path:

1.  **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 ($$a_d$$) is $$a$$
    - Ratio ($$r_d$$) is $$r$$
    - Sum of downward path: $$S_{\text{down}} = \frac{a_d}{1 - r_d} = \frac{a}{1 - r}$$

2.  **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 ($$a_u$$) is $$ar$$
    - Ratio ($$r_u$$) is $$r$$
    - Sum of upward path: $$S_{\text{up}} = \frac{a_u}{1 - r_u} = \frac{ar}{1 - r}$$

Visible text: 1. **Downward Path:** The ball falls from height , then falls again after the first bounce (), falls again after the second bounce (), and so on.

 - Downward series: 
 - First term () is 
 - Ratio () is 
 - Sum of downward path: 

2. **Upward Path:** The ball moves up after the first bounce (), up again after the second bounce (), and so on.
 - Upward series: 
 - First term () is 
 - Ratio () is 
 - Sum of upward path:

Component: MathContainer
Children:

```math
\text{Total Distance} = \text{Sum of Downward Path} + \text{Sum of Upward Path}
```

```math
\text{Total} = S_{\text{down}} + S_{\text{up}} = \frac{a}{1 - r} + \frac{ar}{1 - r}
```

```math
\text{Total} = \frac{a(1 + r)}{1 - r}
```

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

Visible text: For the tennis ball example with and :

```math
\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}
```

So, the total distance traveled by the ball until it stops is $$\frac{5}{3} \text{ meters}$$.

Visible text: So, the total distance traveled by the ball until it stops is .

**Alternative Method (Following the Hint):**
The total distance can also be calculated as: first downward path plus $$2 \text{ times}$$ the sum of all upward paths.

Visible text: **Alternative Method (Following the Hint):**
The total distance can also be calculated as: first downward path plus the sum of all upward paths.

Component: MathContainer
Children:

```math
\text{Total} = a + 2 \times S_{\text{up}} = a + 2 \times \left( \frac{ar}{1 - r} \right)
```

```math
\text{Total} = 1 + 2 \times \left( \frac{1 \times \frac{1}{4}}{1 - \frac{1}{4}} \right) = 1 + 2 \times \left( \frac{1/4}{3/4} \right) = 1 + 2 \times \left( \frac{1}{3} \right) = 1 + \frac{2}{3} = \frac{5}{3}
```

The result is the same!