Sum of an Infinite Geometric Series
When an infinite geometric series converges and how to sum it with a/(1 - r). Includes |r| < 1 test, repeating decimals, and why r >= 1 diverges.
Sum of an Infinite Geometric Series
The most common mistake is thinking that adding infinitely many numbers must produce an infinite result. That is false. A geometric series with a common ratio whose absolute value is less than 1 converges to a finite number. The formula sum = a/(1 - r) gives that number directly, where a is the first term and r is the common ratio. If |r| ≥ 1, the series diverges, and the formula does not apply. OpenStax Calculus Vol. 2 section 5.2 states this convergence condition and the sum formula.
Finite vs. Infinite Geometric Series: What Changes
A finite geometric series has a last term. Its sum is Sn = a(1 − rn)/(1 − r) for r ≠ 1, where n is the number of terms. The infinite geometric series removes the upper bound. A finite sum always exists. An infinite sum exists only if the partial sums approach a limit. The finite formula is the starting point for the infinite one: take the limit of the partial sum as n → ∞.
The Convergence Test: |r| < 1
Before applying the formula a/(1 − r), check the common ratio r. If |r| < 1, the terms shrink fast enough for the sum to converge. If |r| = 1, the series does not converge; the terms either stay constant (r = 1) or alternate without decaying (r = −1). If |r| > 1, the terms grow, and the partial sums blow up.
OpenStax Calculus Vol. 2 section 5.2 gives the geometric series convergence condition as |r| < 1. The same source states divergence for |r| ≥ 1. Skipping the convergence check is the most common error; it leads to applying a/(1 − r) to a divergent series, producing a nonsense number.
The Formula S = a/(1 − r) and Where It Comes From
The formula is the limit of the partial sums. For a finite geometric series, the nth partial sum is Sn = a(1 − rn)/(1 − r). Take the limit as n → ∞. If |r| < 1, then rn → 0, so Sn → a/(1 − r). If |r| ≥ 1, rn does not approach 0, and the limit does not exist.
This derivation is in OpenStax Calculus Vol. 2 section 5.2. The variables are the same: a is the first term, r is the common ratio. The formula only holds when |r| < 1.
Worked Examples: Four Scenarios
Example 1: Simple Positive Ratio
Find the sum of 1/2 + 1/4 + 1/8 + 1/16 + … . First term a = 1/2. Common ratio r = 1/2. |r| = 0.5, which is less than 1, so the series converges. Apply the formula: S = (1/2) / (1 − 1/2) = (1/2) / (1/2) = 1.
Example 2: Alternating Ratio
Find the sum of 1 − 1/3 + 1/9 − 1/27 + … . First term a = 1. Common ratio r = −1/3. |r| = 1/3, less than 1, so converges. S = 1 / (1 − (−1/3)) = 1 / (1 + 1/3) = 1 / (4/3) = 3/4.
Example 3: Divergent Ratio
Find the sum of 5 + 10 + 20 + 40 + … . First term a = 5. Common ratio r = 2. |r| = 2, greater than 1, so the series diverges. Do not apply a/(1 − r). The sum is infinite.
Example 4: Starting Index Not Zero
Find the sum of 3/4 + 3/8 + 3/16 + … . First term a = 3/4. Common ratio r = 1/2. S = (3/4) / (1 − 1/2) = (3/4) / (1/2) = 3/2. The index can start at any number; what matters is the first term and the ratio.
Repeating Decimals as Geometric Series
Every repeating decimal is a geometric series. For 0.999…, write it as 9/10 + 9/100 + 9/1000 + … . First term a = 9/10, common ratio r = 1/10. |r| < 1, so S = (9/10) / (1 − 1/10) = (9/10) / (9/10) = 1. The infinite geometric series sum shows that 0.999… equals 1 exactly.
For 0.272727…, write it as 27/100 + 27/10000 + … . First term a = 27/100, r = 1/100. S = (27/100) / (99/100) = 27/99 = 3/11. This method converts any repeating decimal to a fraction without memorising patterns.
Partial-Sum Convergence Table
The table below shows how partial sums of the series 1/2 + 1/4 + 1/8 + … approach the limit 1 as you add more terms. The first column is the number of terms added. The second column is the partial sum value. The third column is the difference from the limit.
| Number of Terms | Partial Sum S_n | Difference from Limit (1) |
|---|---|---|
| 1 | 0.5 | 0.5 |
| 2 | 0.75 | 0.25 |
| 3 | 0.875 | 0.125 |
| 4 | 0.9375 | 0.0625 |
| 5 | 0.96875 | 0.03125 |
| 10 | 0.9990234375 | 0.0009765625 |
| 15 | 0.99993896484375 | 0.00006103515625 |
What a Calculator Does: Partial Sums as an Approximation
A calculator cannot add infinitely many terms. When you use a TI-84 Plus CE Σ( template or sum(/seq( method, it adds a finite number of terms. The result is the nth partial sum, not the infinite sum. For a convergent geometric series with |r| < 1, that partial sum gets close to the limit, but it is never equal unless the calculator uses the formula directly.
It is not the limit; it is an approximation. The formula a/(1 − r) gives the exact limit.
Other Infinite Series People Confuse With Geometric
An infinite geometric series is not the only kind of infinite series. Two common ones that people mistake for geometric are the harmonic series and p-series.
The harmonic series is 1 + 1/2 + 1/3 + 1/4 + … . It has no constant ratio between terms, so it is not geometric. It diverges, even though its terms approach zero. The p-series Σ 1/np converges only when p > 1. For p = 1, it is the harmonic series and diverges. For p = 2, it converges, but it is still not geometric because there is no common ratio. If you try to apply a/(1 − r) to either, you will get a wrong answer.
Common Questions
Can I use a/(1 − r) if the series does not start at n = 1?
Yes. Identify the first term of the series, whatever it is, and use that as a. The index starting point does not change the formula. For example, the series 3/8 + 3/16 + 3/32 + … has a = 3/8, r = 1/2, and sum = (3/8)/(1/2) = 3/4.
What happens if r = 1?
If r = 1, every term equals a, so the series is a + a + a + … . The nth partial sum is n·a, which grows without bound as n increases. The series diverges. The formula a/(1 − r) is undefined because it involves division by zero, and the partial sum formula a(1 − r^n)/(1 − r) is also undefined for r = 1.
Does the sum formula work for negative values of r?
Yes, as long as |r| < 1. A negative r produces an alternating series. For example, the series 1 − 1/2 + 1/4 − 1/8 + … has a = 1, r = −1/2. |r| = 1/2, so it converges. S = 1/(1 − (−1/2)) = 2/3.
Why does the formula use a/(1 − r) and not a/(r − 1)?
The derivation from the partial sum gives a(1 − r^n)/(1 − r). Taking the limit as n → ∞ with |r| < 1 makes r^n → 0, leaving a/(1 − r). The denominator is (1 − r), not (r − 1). Using (r − 1) would flip the sign and give the wrong answer.
Can I convert any repeating decimal to a fraction using geometric series?
Yes, if the decimal is purely repeating. Write the repeating part as a geometric series. For 0.333…, the series is 3/10 + 3/100 + 3/1000 + … . First term a = 3/10, r = 1/10. S = (3/10)/(9/10) = 1/3. For mixed repeating decimals, you need to separate the non-repeating part first.
What if the common ratio is a fraction like 2/3? Does the formula still work?
Yes. The formula works for any real number r as long as |r| < 1, including fractions. For example, the series 1 + 2/3 + 4/9 + 8/27 + … has a = 1, r = 2/3. |r| = 2/3 < 1. S = 1/(1 − 2/3) = 1/(1/3) = 3.
Does the geometric series converge if r is exactly 0?
Yes. If r = 0, the series is a + 0 + 0 + 0 + … . The first term is a, and all subsequent terms are zero. The partial sums are all a after the first term, so the series converges to a. The formula a/(1 − 0) = a gives the correct result.