Sum of an Infinite GP - Formula & Examples

What Is The Sum Of An Infinite GP?

The sum of an infinite GP is the finite value that the running total of an unending geometric progression approaches, provided the terms shrink fast enough. For the progression $a, ar, ar^2, ar^3, \dots$ continuing forever, the sum is

$$S_\infty = \frac{a}{1 - r}, \qquad |r| < 1$$

The condition $|r| < 1$ is not decoration. It is the entire difference between a series that settles on a number and one that runs off to infinity. A geometric progression is a sequence where each term is the previous one times a fixed ratio $r$; you can review the basics in the geometric sequence explainer. When $|r| < 1$, each term is a fraction of the last, so the terms melt toward zero.

Why Does The Series Converge Only When |r| < 1?

This is the question that actually matters, and it has a clean answer. Start from the finite sum of a GP, covered in full in the sum of a GP article:

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

Now ask what happens to $r^n$ as $n$ grows without bound.

• If $|r| < 1$, raising a fraction to higher and higher powers drives it toward zero. For example $\left(\frac{1}{2}\right)^{10} = \frac{1}{1024}$, and it only gets tinier. So $r^n \to 0$.

• If $|r| \geq 1$, the terms do not shrink. They stay the same size ($r = 1$) or grow ($|r| > 1$), so the running total never settles.

Substitute $r^n \to 0$ into the finite formula:

$$S_\infty = \frac{a(1 - 0)}{1 - r} = \frac{a}{1 - r}$$

That is the whole derivation. The infinite formula is the finite formula with its one shrinking piece sent to zero. The condition $|r| < 1$ is just the requirement that the piece can shrink to zero.

Variable Glossary:

What Happens When |r| ≥ 1?

The series diverges - it has no finite sum. If $r = 2$, the progression $3, 6, 12, 24, \dots$ keeps growing and the total races to infinity; if $r = 1$, every term equals $a$, and adding $a$ forever also gives infinity (unless $a = 0$); if $r = -1$, the partial sums bounce between two values and never settle. In all of these, the formula $\frac{a}{1-r}$ gives a nonsense answer or no answer, so you must not apply it. Checking $|r| < 1$ before reaching for the formula is the single non-negotiable step.

Examples Of The Sum Of An Infinite GP

Example 1

Find the sum of $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$.

First term $a = 1$, ratio $r = \frac{1}{2}$. Since $|r| < 1$, a sum exists:

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

$$S_\infty = \frac{1}{\frac{1}{2}}$$

$$S_\infty = 2$$

Final answer: $S_\infty = 2$. This is the walk-to-the-wall problem made exact: infinitely many halving steps cover exactly $2$ units.

Example 2

Find the sum of $\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots$.

Here $a = \frac{1}{3}$ and $r = \frac{1}{3}$:

$$S_\infty = \frac{\frac{1}{3}}{1 - \frac{1}{3}}$$

$$S_\infty = \frac{\frac{1}{3}}{\frac{2}{3}}$$

$$S_\infty = \frac{1}{3} \times \frac{3}{2} = \frac{1}{2}$$

Final answer: $S_\infty = \dfrac{1}{2}$.

Example 3

Find the sum of $3 + 6 + 12 + 24 + \cdots$.

The instinct is to plug in $a = 3$ and start dividing. Try it:

$$S_\infty = \frac{3}{1 - 2} = \frac{3}{-1} = -3$$

Take a second. The series adds only positive, growing numbers, so a total of $-3$ is impossible - a sum smaller than the first term cannot be right. The error was applying the formula without checking the ratio: here $r = \frac{6}{3} = 2$, and $|r| = 2 \geq 1$, so the series diverges and there is no finite sum. The negative answer was the formula's way of warning that it was being used outside its domain.

Final answer: the series diverges; no sum exists. Always test $|r| < 1$ first.

Example 4

A bouncing ball. A ball is dropped from $10$ metres. Each bounce reaches $\frac{3}{5}$ of the previous height. Find the total vertical distance travelled before it comes to rest.

The ball falls $10$ m, then for each bounce it goes up and down the same height. The up-distances form a GP with $a = 10 \times \frac{3}{5} = 6$ and $r = \frac{3}{5}$; the down-distances match. So total distance is the first drop plus twice the sum of the bounce heights:

$$\text{up-bounces sum} = \frac{6}{1 - \frac{3}{5}} = \frac{6}{\frac{2}{5}} = 15$$

$$\text{total} = 10 + 2 \times 15 = 40 \text{ m}$$

Final answer: $40$ metres. The ball bounces infinitely often in principle, yet travels a finite distance - the same convergence idea, dressed as physics.

Example 5

Write the repeating decimal $0.444\ldots$ as a fraction.

A repeating decimal is an infinite GP in disguise:

$$0.444\ldots = \frac{4}{10} + \frac{4}{100} + \frac{4}{1000} + \cdots$$

This is a GP with $a = \frac{4}{10}$ and $r = \frac{1}{10}$:

$$S_\infty = \frac{\frac{4}{10}}{1 - \frac{1}{10}}$$

$$S_\infty = \frac{\frac{4}{10}}{\frac{9}{10}}$$

$$S_\infty = \frac{4}{9}$$

Final answer: $0.444\ldots = \dfrac{4}{9}$.

Example 6

Show that $0.999\ldots = 1$.

This is the example that starts arguments, and the infinite GP settles it. Write it as a series:

$$0.999\ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots$$

A GP with $a = \frac{9}{10}$, $r = \frac{1}{10}$:

$$S_\infty = \frac{\frac{9}{10}}{1 - \frac{1}{10}}$$

$$S_\infty = \frac{\frac{9}{10}}{\frac{9}{10}}$$

$$S_\infty = 1$$

Final answer: $0.999\ldots = 1$ exactly. Not "approximately", not "rounds to". The infinite GP gives the precise value, which is why mathematicians treat the two as the same number. The friction students feel here is real - the idea that an endless string of $9$s equals $1$ rather than falling just short took the field itself a long time to make rigorous.

Why The Convergence Condition Is The Real Lesson

"Each term is a fraction of the last, so the leftover keeps halving."

The formula $\frac{a}{1-r}$ is easy to memorise. The condition $|r| < 1$ is the part that carries the understanding, and it is the part students drop first. The whole reason an infinite sum can be finite is that the terms vanish fast enough; the moment they stop vanishing, the sum is gone.

This is also a doorway to a much larger subject. Infinite GPs are the simplest convergent series, the first ones students meet before calculus generalises the idea to every kind of infinite sum. The distinction between a series that converges and one that diverges - decided here by a single inequality - becomes one of the central questions of analysis. Zeno's paradox of Achilles and the tortoise dissolves the instant you recognise the gap distances as a convergent GP: the paradox was never about motion, but about not yet having this formula.

Tripping Points To Avoid

Mistake 1: Applying the formula without checking |r|

Where it slips in: Any problem where the ratio is greater than or equal to $1$ in size.

Don't do this: Plug $a$ and $r$ straight into $\frac{a}{1-r}$ and report a number, even when the terms are clearly growing.

The correct way: Test $|r| < 1$ before anything else. If the test fails, write "diverges - no sum" and stop. The habit of checking the ratio first is what stops a student from reporting a finite total for a series that runs to infinity, which is the single most common error on this topic.

Mistake 2: Confusing the infinite sum with the finite sum

Where it slips in: A problem says "find the sum to infinity" but the student computes a partial sum $S_n$ instead, or vice versa.

Don't do this: Use $S_\infty = \frac{a}{1-r}$ when the question asks for the first $8$ terms only.

The correct way: Match the formula to the question. "Sum to infinity" or "the series" means $S_\infty = \frac{a}{1-r}$; "sum of the first $n$ terms" means the finite $S_n = \frac{a(1-r^n)}{1-r}$. The two share a denominator, which is exactly why the memorizer who relies on shape rather than meaning confuses them.

Mistake 3: Mishandling r in a repeating decimal

Where it slips in: Turning $0.272727\ldots$ into a fraction.

Don't do this: Set $r = \frac{1}{10}$ when the repeating block is two digits long.

The correct way: The ratio depends on the length of the repeating block. For a two-digit block like $27$, the terms are $\frac{27}{100}, \frac{27}{10000}, \dots$, so $a = \frac{27}{100}$ and $r = \frac{1}{100}$. Count the digits in the repeating block to set $r$ correctly.

Practice Questions on the Sum of an Infinite GP

Test $|r| < 1$ before applying the formula. Answers follow.

1. Find the sum of $4 + 2 + 1 + \frac{1}{2} + \cdots$.

2. Find the sum of $9 + \frac{9}{4} + \frac{9}{16} + \cdots$.

3. Find the sum of $1 + 3 + 9 + 27 + \cdots$, or state that it diverges.

4. Find the sum of $\frac{2}{3} - \frac{2}{9} + \frac{2}{27} - \cdots$.

5. Write the repeating decimal $0.515151\ldots$ as a fraction using an infinite GP.

Answers:

1. $a = 4$, $r = \frac{1}{2}$: $S_\infty = \dfrac{4}{1 - \frac{1}{2}} = 8$.

2. $a = 9$, $r = \frac{1}{4}$: $S_\infty = \dfrac{9}{1 - \frac{1}{4}} = 12$.

3. $r = 3$, so $|r| \geq 1$: the series diverges and has no finite sum.

4. $a = \frac{2}{3}$, $r = -\frac{1}{3}$: $S_\infty = \dfrac{\frac{2}{3}}{1 - \left(-\frac{1}{3}\right)} = \dfrac{\frac{2}{3}}{\frac{4}{3}} = \dfrac{1}{2}$.

5. The repeating block is two digits, so $a = \frac{51}{100}$, $r = \frac{1}{100}$: $S_\infty = \dfrac{\frac{51}{100}}{1 - \frac{1}{100}} = \dfrac{51}{99} = \dfrac{17}{33}$.

Key Takeaways

• The sum of an infinite GP is $S_\infty = \frac{a}{1-r}$, valid only when $|r| < 1$.

• The formula is the finite GP sum with $r^n$ driven to zero as the terms vanish.

• If $|r| \geq 1$, the series diverges and has no finite sum - check this first.

• Repeating decimals are infinite GPs; $0.999\ldots$ equals exactly $1$.

• Convergence here is the first example of the broader idea that some infinite sums settle on a number and others do not.

A Practical Next Step

Now turn $0.666\ldots$ into a fraction using the formula, then check your ratio by counting the digits in the repeating block. If the convergence condition still feels abstract, list the first six partial sums of $1 + \frac{1}{2} + \frac{1}{4} + \cdots$ by hand and watch them inch toward $2$

At Bhanzu, our trainers anchor infinite series to the walking-to-the-wall picture before any formula appears, so the convergence condition feels inevitable rather than memorised. Want a live trainer to explore more infinite-series problems? Book a free demo class.