Geometric Series: When They Converge and What They Sum To

A geometric series is the one series you can actually sum exactly. Identify it by the fixed ratio between consecutive terms. If you can write the series as Σ a·rⁿ, starting at n = 0, then a is the first term and r is the common ratio. For example, 2 + 4 + 8 + 16 + ... gives a = 2 and r = 2. The series 100 + 10 + 1 + 0.1 + ... gives a = 100 and r = 0.1. If the terms are not multiplied by a fixed number each step, it is not a geometric series. That distinction matters: many students confuse a geometric series with a p-series. A p-series has terms 1/n^p; the ratio between terms is not fixed, and no constant r exists. Check the ratio: divide term n+1 by term n. If that quotient is the same for every n, you have a geometric series. If it changes, move on to another test.

A geometric series converges if and only if the absolute value of the common ratio is less than 1. That is the sole condition: |r| < 1. If |r| ≥ 1, the series diverges. This is not a test that has an inconclusive result. It gives a binary answer every time. Compare that to the ratio test, which becomes inconclusive when the limit equals 1. The geometric series test does not have that problem. For |r| < 1, the partial sums approach a finite number. For r = 1, the series is just a repeated a, and it diverges because the terms do not approach 0. For r = -1, the series alternates between a and -a and diverges because the partial sums never settle. For |r| > 1, the terms grow in magnitude and the series diverges by the divergence test. The convergence condition is the most reliable rule in series testing.

When the geometric series converges, its sum is a / (1 - r), but only if the first term is a and the series starts at n = 0. If the series starts at n = 1, the sum formula changes to a·r / (1 - r). The difference is the single most common error students make. Stewart, Calculus, section 11.2, gives the example Σ (1/2)ⁿ = 2 when the series starts at n = 0. The same series starting at n = 1 sums to 1. The formula produces a / (1 - r) = 1 / (1 - 1/2) = 2, but only for n = 0. For n = 1, the first term is 1/2, so the sum is (1/2) / (1 - 1/2) = 1. Always check the index. The partial sum formula S_n = a(1 - rⁿ) / (1 - r) works for finite sums regardless of index, but the infinite formula requires matching the first term to a. Write the series out for the first three terms, identify a, then apply the formula.

Example 1: Standard Geometric Series

Consider Σ 5·(0.3)ⁿ starting at n = 0. Here a = 5 and r = 0.3. Since |0.3| < 1, the series converges. The sum is a / (1 - r) = 5 / (1 - 0.3) = 5 / 0.7 ≈ 7.14.

Example 2: Index Shift

Take Σ 3·(0.4)ⁿ starting at n = 1. Write the first term: n = 1 gives 3·(0.4) = 1.2. So a = 1.2, r = 0.4. Sum = a / (1 - r) = 1.2 / (1 - 0.4) = 1.2 / 0.6 = 2. If you mistakenly used a = 3, you would get 3 / 0.6 = 5, which is wrong.

Example 3: Repeating Decimal

The repeating decimal 0.333... is a geometric series. Write it as 0.3 + 0.03 + 0.003 + ... = 3/10 + 3/100 + 3/1000 + ... This is Σ (3/10)·(1/10)ⁿ starting at n = 0. Then a = 3/10, r = 1/10. Sum = (3/10) / (1 - 1/10) = (3/10) / (9/10) = 1/3. The same method works for any repeating decimal: break it into a geometric series and sum it.

Example 4: Divergent Series

Consider Σ 2·(1.5)ⁿ starting at n = 0. Here r = 1.5. Since |1.5| > 1, the series diverges. No sum exists. The terms grow without bound, and the partial sums do not approach a finite limit.

Example 5: Negative Ratio

Stewart gives Σ (-1/2)ⁿ = 2/3 for n starting at 0. Here a = 1, r = -1/2. Sum = 1 / (1 - (-1/2)) = 1 / (1 + 1/2) = 1 / (3/2) = 2/3. The series alternates but converges because |r| < 1.

Because the geometric series test is decisive and its sum is known, geometric series serve as the primary benchmark for the comparison test and the limit comparison test. When you encounter a series that looks like a geometric series but is not exactly one, you compare it to a convergent or divergent geometric series. For example, the series Σ 2ⁿ / (3ⁿ + 1) behaves like Σ (2/3)ⁿ for large n. Since Σ (2/3)ⁿ is a convergent geometric series (r = 2/3), the comparison test or limit comparison test can confirm convergence. The geometric series is the simplest known series to use as a baseline. Without it, the comparison test would have no anchor. The p-series is the other standard benchmark, but the geometric series handles exponential and factorial terms better. When you are testing a series and need a known convergent or divergent series, the geometric series is the first one to try.

Index Confusion

The sum formula changes with the starting index. Write out the first two terms before applying a / (1 - r).

Treating Non-Geometric Series as Geometric

A series like Σ 2ⁿ / n is not geometric. The ratio between terms is not fixed because of the n in the denominator. Check the ratio explicitly.

Forgetting the Absolute Value

The geometric series test uses |r|, not r. A ratio of -0.9 gives |r| = 0.9, so the series converges. A ratio of -1.1 gives |r| = 1.1, so the series diverges.

Misapplying the Divergence Test

If a geometric series has |r| < 1, the limit of a_n is 0, but that does not prove convergence. The divergence test only proves divergence when the limit is not 0. You need the geometric series test for a positive result.

Condition	Result	Sum Formula
|r| < 1	Converges	a/(1-r) for n=0; a·r/(1-r) for n=1
|r| = 1 (r=1)	Diverges	No sum
|r| = 1 (r=-1)	Diverges	No sum
|r| > 1	Diverges	No sum

If you see a series that has terms multiplied by a constant factor each step, apply the geometric series test immediately. It is the fastest test with the clearest answer. The ratio test works on many series, but the geometric series test is simpler when it applies. For a series like Σ 5ⁿ / 2ⁿ⁺¹, rewrite it as (1/2) Σ (5/2)ⁿ, then apply the geometric series test. Since |5/2| > 1, the series diverges. That takes ten seconds. Do not overcomplicate it. Use the geometric series test before the ratio test, before the comparison test, before the integral test. It is the first test in the strategy for testing series because it is decisive and simple. The one thing that most often goes wrong: forgetting to check the starting index. Write the series out for n = 0 and n = 1, find a, then apply the formula. That single step will save you from the most common error.