Geometric Series and Finite Geometric Sums

A geometric series is the sum of the terms of a geometric sequence, and a finite geometric sum with n terms equals a₁(1 − rⁿ)/(1 − r) for r ≠ 1. In brief: 2 + 6 + 18 + 54 + 162 has a₁ = 2, r = 3 and n = 5, so the sum is 2(3⁵ − 1)/(3 − 1) = 242. The sum of the infinite series a₁ + a₁r + a₁r² + … is a₁/(1 − r) if |r| < 1 and does not exist as a number otherwise.

Geometric sums are among the most useful formulas in mathematics. They give the future value of regular savings, the total distance of a bouncing ball, the repeating decimals as fractions and the number of nodes in a tree. This guide derives the finite sum, shows several forms of it, explains the infinite geometric series and its condition for convergence, and covers applications, with a table of partial sums.

What is the finite geometric sum?

For a geometric sequence a_k = a₁r^k−1, the partial sum is S_n = a₁ + a₁r + … + a₁rⁿ⁻¹. For r ≠ 1:

S_n = a₁(1 − rⁿ)/(1 − r).

For r = 1 all terms equal a₁, so S_n = n a₁. The derivation, by multiplying the sum by r and subtracting. A closely related factorization is 1 − rⁿ = (1 − r)(1 + r + r² + … + rⁿ⁻¹), for instance x³ − 1 = (x − 1)(x² + x + 1).

Which forms of the formula are useful?

• For |r| < 1: S_n = a₁(1 − rⁿ)/(1 − r).
• For |r| > 1: S_n = a₁(rⁿ − 1)/(r − 1), which avoids negative numbers.
• With the last term a_n = a₁rⁿ⁻¹: S_n = (a₁ − a_nr)/(1 − r).
• Sum starting from k = 0: 1 + r + … + rⁿ = (rⁿ⁺¹ − 1)/(r − 1).

Example 1

3 + 6 + 12 + … + 768. The ratio is 2 and the last term 768 = 3 · 2⁸, so there are 9 terms. S = 3(2⁹ − 1) = 1,533. With the last-term form: (768 · 2 − 3)/(2 − 1) = 1,533 ✓.

Example 2: a decreasing series

8 + 4 + 2 + 1 + 1/2 + … (8 terms): r = 1/2, S₈ = 8(1 − 1/256)/(1/2) = 16 · 255/256 = 15.9375.

Example 3: with a negative ratio

1 − 3 + 9 − 27 + 81 − 243 (6 terms): r = −3, S = (1 − 729)/(1 + 3) = −728/4 = −182. Check: 1 − 3 = −2, +9 = 7, −27 = −20, +81 = 61, −243 = −182 ✓.

When does an infinite geometric series converge?

The partial sums are S_n = a₁(1 − rⁿ)/(1 − r). If |r| < 1, then rⁿ → 0 as n → ∞, so S_n approaches a₁/(1 − r). Therefore

a₁ + a₁r + a₁r² + … = a₁/(1 − r), for |r| < 1.

If |r| ≥ 1 (and a₁ ≠ 0), the terms do not tend to 0, and the series diverges. For example, 1 + 2 + 4 + … grows without bound and 1 − 1 + 1 − 1 + … oscillates between 1 and 0.

Example 4

1 + 1/2 + 1/4 + 1/8 + … = 1/(1 − 1/2) = 2. The partial sums 1, 1.5, 1.75, 1.875, … approach 2.

Example 5: with alternating signs

6 − 2 + 2/3 − 2/9 + … has r = −1/3, so the sum is 6/(1 + 1/3) = 4.5.

Example 6: repeating decimals

0.272727… = 0.27 + 0.0027 + 0.000027 + … = 0.27/(1 − 0.01) = 27/99 = 3/11. And 0.999… = 0.9/(1 − 0.1) = 1, so 0.999… equals 1.

Example 7: a convergence condition

For which x does 1 + x + x² + … converge? For |x| < 1, and then the sum is 1/(1 − x). For x = 0.2: 1.25. For the series with terms (x − 1)ⁿ: converges for |x − 1| < 1, that is 0 < x < 2.

How do you use geometric sums?

Example 8: a bouncing ball

A ball is dropped from 10 m and rebounds to 60% of the previous height each time. The total distance is 10 + 2 · 6 + 2 · 3.6 + … = 10 + 2 · 6/(1 − 0.6) = 10 + 30 = 40 m.

Example 9: regular savings (annuity)

Depositing 500 dollars at the end of each year for 20 years at 3% yearly: the future value is 500(1 + 1.03 + … + 1.03¹⁹) = 500 · (1.03²⁰ − 1)/0.03 = 500 · 26.870 = 13,435.19 dollars.

Example 10: a loan

A loan of 10,000 dollars at 6% yearly is repaid by equal yearly payments P over 5 years. The present value of the payments is P(1 − 1.06⁻⁵)/0.06 = 10,000, so P = 10000 · 0.06/(1 − 0.7473) = 2,373.96 dollars.

Example 11: a tree

A binary tree with 12 levels, where level k has 2^k−1 nodes, has 2¹² − 1 = 4,095 nodes.

Example 12: the Koch snowflake perimeter

A construction multiplies the perimeter by 4/3 at each step, so the perimeter grows without bound, since 4/3 > 1. But the added area forms a series with the ratio 4/9, which converges, so the area is finite.

Partial sums table

The left column approaches 2, the sum of the infinite series. The right column grows without bound.

Where do geometric series come from?

Archimedes summed the geometric series with the ratio 1/4 in his computation of the area of a parabolic segment, around 250 BCE: 1 + 1/4 + 1/16 + … = 4/3. Zeno’s paradoxes of motion, such as Achilles and the tortoise, are about infinite geometric series with a finite sum. The modern theory of convergence developed in the seventeenth to nineteenth centuries.

Where are they used today?

Banks use geometric sums for annuities, mortgages and bond prices. In economics, the multiplier effect, in which spending is re-spent with a fraction c each round, gives a total effect 1/(1 − c) times the original: for c = 0.8, 5 times. Signal processing and probability use geometric series to compute expected values, such as the expected number of tosses until the first head.

Which mistakes are common?

Using the infinite sum when |r| ≥ 1

The formula a₁/(1 − r) needs |r| < 1. For r = 2, the series diverges.

Wrong number of terms

Count terms with n = log_r(a_n/a₁) + 1. A sum from k = 0 to n has n + 1 terms.

Using a₁ as the ratio

The ratio is the quotient of consecutive terms, not the first term.

Mixing up the first term

In 5 + 5r + 5r² + …, a₁ = 5. For a series starting at 5r, the first term is 5r.