The Limiting Sum of a Geometric Series (HSC SSCE Mathematics Advanced): Revision Notes

The Limiting Sum of a Geometric Series

Understanding limiting sums through an example

Let's explore the concept of a limiting sum using an interesting scenario. Imagine a frog that's $8$ metres away from a waterhole. The frog is dying of thirst and starts jumping toward the water. On its first jump, it covers $4$ metres. On the second jump, it covers $2$ metres. Each subsequent jump is half the distance of the previous jump.

The question is: will the frog ever reach the water?

The frog's jumps form a geometric progression (GP). We can track both the individual jump distances ($T_n$) and the cumulative distance covered ($S_n$):

This means the frog's total distance gets closer and closer to $8$ metres. So even though each individual jump gets smaller, the frog will eventually get close enough to the waterhole. If the frog can extend its tongue even a tiny fraction of a millimetre, it will reach the water and be saved!

Analyzing the limiting sum

Let's examine this more carefully using the formula for the sum of a geometric series. We need to look at what happens to the sum $S_n$ as $n$ becomes very large (as $n \to \infty$).

The series $4 + 2 + 1 + \frac{1}{2} + \cdots$ is a GP with first term $a = 4$ and common ratio $r = \frac{1}{2}$.

Using the formula for the sum of the first $n$ terms:

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

Since $r < 1$, we use this version of the formula.

Substituting our values:

$S_n = \frac{4(1-(\frac{1}{2})^n)}{1-\frac{1}{2}}$

$= 4 \times (1-(\frac{1}{2})^n) \div \frac{1}{2}$

$= 8(1-(\frac{1}{2})^n)$

Now, what happens to $(\frac{1}{2})^n$ as $n$ increases?

$(\frac{1}{2})^2 = \frac{1}{4}, \quad (\frac{1}{2})^3 = \frac{1}{8}, \quad (\frac{1}{2})^4 = \frac{1}{16}, \quad (\frac{1}{2})^5 = \frac{1}{32}, \quad (\frac{1}{2})^6 = \frac{1}{64}, \ldots$

Therefore:

$S_n = 8(1-(\frac{1}{2})^n) \to 8(1-0) = 8$

This confirms our observation from the table: the limiting sum is $\mathbf{8}$.

Different ways to express limiting sums

There are several equivalent ways to describe what we've just discovered. Using the series $4 + 2 + 1 + \frac{1}{2} + \cdots$ as our example:

Five equivalent statements:

• $S_n \to 8$ as $n \to \infty$

  This reads as: "$S_n$ has limit $8$ as $n$ increases without bound"

• $\lim_{n \to \infty} S_n = 8$

  This reads as: "The limit of $S_n$, as $n$ increases without bound, is $8$"

• The series $4 + 2 + 1 + \frac{1}{2} + \cdots$ has limiting sum $S_{\infty} = 8$

• The series $4 + 2 + 1 + \frac{1}{2} + \cdots$ converges to the limit $S_{\infty} = 8$

• $4 + 2 + 1 + \frac{1}{2} + \cdots = 8$

The general formula

Now let's develop a general formula that works for any geometric series.

Consider a GP with first term $a$ and common ratio $r$, so that:

$T_n = ar^{n-1} \quad \text{and} \quad S_n = \frac{a(1-r^n)}{1-r}$

Suppose that the ratio $r$ lies in the interval $-1 < r < 1$ (equivalently, $|r| < 1$).

As $n$ increases, the successive powers $r^1, r^2, r^3, r^4, \ldots$ get smaller and smaller in absolute value.

This means:

• $r^n \to 0$ as $n \to \infty$
• $1 - r^n \to 1$ as $n \to \infty$

Therefore, both the $n$th term $T_n$ and the partial sum $S_n$ converge to limits:

$\lim_{n \to \infty} T_n = \lim_{n \to \infty} ar^{n-1} = 0$

$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{a(1-r^n)}{1-r} = \frac{a(1-0)}{1-r} = \frac{a}{1-r}$

Key point: A geometric series only has a limiting sum when the absolute value of the common ratio is less than $1$.

Worked examples

Sigma notation for infinite sums

When $-1 < r < 1$ and the GP converges, we can write the limiting sum using sigma notation or using dots:

$\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}$

or equivalently:

$a + ar + ar^2 + \cdots = \frac{a}{1-r}$

Remember!