Infinite Geometric Series

Infinite Geometric Series — Moosa Academy

Adding endlessly many numbers sounds like it must give an endlessly large answer. Sometimes it does. But when each term is a small enough fraction of the one before, the running total closes in on a fixed value and stops there — and a single condition on the ratio tells you which of the two will happen.

Concept Convergent and divergent

Halve a square, then halve what is left, and keep going. Infinitely many pieces are created, yet together they never exceed the original square. That is a convergent series: the partial sums approach a definite value.

A divergent series has no such limit — its partial sums grow without bound and no total exists.

Theorem The sum to infinity

An infinite geometric series converges when $|r| < 1$ , and its sum is $S = \dfrac{a_1}{1 - r}$ . When $|r| \geq 1$ the series diverges and has no sum.

The condition is on the absolute value, so $r = -0.9$ converges just as $r = 0.9$ does. What matters is that each term is smaller in size than the last, no matter which side of zero it sits on.

Example Convergent or divergent?

a) $54 + 36 + 24 + \cdots$

Find the ratio:

r = \dfrac{36}{54} = \dfrac{2}{3}

Check:

\left|\dfrac{2}{3}\right| < 1

Convergent — a sum exists

b) $8 + 12 + 18 + \cdots$

Find the ratio:

r = \dfrac{12}{8} = 1.5

Check:

|1.5| > 1

Divergent — no sum exists

Deciding takes one division. Everything else about the series is irrelevant until this test is passed.

Example Finding the sum

a) $54 + 36 + 24 + \cdots$

a_1 = 54

and

r = \dfrac{2}{3}

, so the series converges

Substitute:

S = \dfrac{54}{1 - \frac{2}{3}}

Simplify the denominator:

1 - \dfrac{2}{3} = \dfrac{1}{3}

Divide:

S = 54 \div \dfrac{1}{3} = 54 \times 3

$S = 162$

b) $6 + 9 + 13.5 + 20.25 + \cdots$

Find the ratio:

r = \dfrac{9}{6} = 1.5

Since

|1.5| > 1

, the terms keep growing

Divergent — there is no sum to find

c) $\displaystyle\sum_{k=1}^{\infty} 12\left(\dfrac{1}{3}\right)^{k-1}$

Read off

a_1 = 12

and

r = \dfrac{1}{3}

Substitute:

S = \dfrac{12}{1 - \frac{1}{3}} = \dfrac{12}{\frac{2}{3}}

Divide:

S = 12 \times \dfrac{3}{2}

$S = 18$

Part (b) is a reminder to test before computing. Feeding a divergent series into the formula produces a number, but that number is meaningless.

Note The two cases side by side

Property	Convergent	Divergent
Condition on $r$	$\|r\| < 1$	$\|r\| \geq 1$
Sum	$S = \dfrac{a_1}{1-r}$	None
Behaviour of $S_n$	Settles toward a fixed value	Grows without bound
Example	$54 + 36 + 24 + \cdots$	$8 + 12 + 18 + \cdots$

Summary

A convergent series has partial sums that approach a fixed value; a divergent one does not.
An infinite geometric series converges exactly when $|r| < 1$ .
For a convergent series, $S = \dfrac{a_1}{1 - r}$ .
When $|r| \geq 1$ no sum exists, and the formula must not be used.
The test uses $|r|$ , so negative ratios such as $-0.9$ converge too.
In sigma notation, read $a_1$ and $r$ straight off the expression and apply the same formula.