Find the sum of the infinite geometric series with a = 1, r = 1/2.

Geometric Series: Sum Formula & Convergence

01 — What Is a Geometric Series?

A geometric series is obtained by adding the terms of a geometric sequence, a sequence in which each term is obtained by multiplying the previous one by a fixed constant called the common ratio r.

The general form looks like this:

General Geometric Series

a + ar + ar² + ar³ + ⋯ = Σ_n=0^∞ arⁿ

where a = first term (a ≠ 0) · r = common ratio

The ratio r is found by dividing any term by the one before it: r = aₙ₊₁ / aₙ.

Some concrete examples to fix the idea:

r = 1/2: 1 + 1/2 + 1/4 + 1/8 + 1/16 + ⋯
r = 3: 2 + 6 + 18 + 54 + 162 + ⋯
r = −1/3: 1 − 1/3 + 1/9 − 1/27 + ⋯ (alternating signs)
r = 1: 5 + 5 + 5 + 5 + ⋯ (constant, diverges)

Geometric series sit at the crossroads of algebra and calculus. They are the simplest infinite series to sum, and the formulas for Taylor series, power series, and even compound interest all trace their roots here.

02 — The Partial Sum Formula

The n-th partial sum Sₙ is the sum of the first n terms (starting from n = 0, so there are n + 1 terms):

Sₙ = a + ar + ar² + ⋯ + arⁿ

Deriving the Formula

Multiply both sides by r:

Sₙ = a + ar + ar² + ⋯ + arⁿ
r · Sₙ = ar + ar² + ⋯ + arⁿ + arⁿ⁺¹

Subtract the second line from the first. All middle terms cancel (this is called a telescoping subtraction):

Sₙ − r·Sₙ = a − arⁿ⁺¹
Sₙ(1 − r) = a(1 − rⁿ⁺¹)

Partial Sum of a Geometric Series (n+1 terms, r ≠ 1)

Sₙ = a · (1 − rⁿ⁺¹) / (1 − r)

Equivalently: Sₙ = a · (rⁿ⁺¹ − 1) / (r − 1) [multiply top & bottom by −1]

When r = 1 the formula breaks down (division by zero); in that case Sₙ = a(n+1) since every term equals a.

03 — Infinite Sum and Convergence

What happens to Sₙ as n → ∞? It depends entirely on r:

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Convergence Condition:
The infinite geometric series Σ arⁿ converges if and only if |r| < 1.
When |r| < 1, the sum is S = a / (1 − r).
When |r| ≥ 1, the series diverges (the partial sums grow without bound or oscillate).

Proof of the Infinite Sum Formula

Start from the partial sum Sₙ = a(1 − rⁿ⁺¹)/(1 − r). Take the limit as n → ∞:

If |r| < 1, then rⁿ⁺¹ → 0 as n → ∞
∴ S = lim Sₙ = a(1 − 0)/(1 − r) = a / (1 − r)

If |r| > 1, then |rⁿ⁺¹| → ∞, so Sₙ diverges. If r = 1, every term equals a ≠ 0, so the sum clearly diverges. If r = −1, the partial sums oscillate between 0 and a, with no limit.

Infinite Geometric Series Sum (|r| < 1)

S∞ = a / (1 − r)

This is the most important formula in the topic. Commit it to memory — it appears in power series, probability generating functions, finance, and physics.

04 — Interactive Convergence Visualiser

Partial sums Sₙ approaching S∞: adjust a and r

First term a: Common ratio r: 0.50

The gold horizontal line is S∞ = a/(1−r). Each bar is a partial sum Sₙ. Watch it converge as n grows.

05 — Partial Sums Table: Watching Convergence in Action

For a = 1 and r = 1/2, the partial sums creep towards S∞ = 2:

n	Term arⁿ	Partial sum Sₙ	Distance from S∞ = 2
0	1	1	1.000000
1	0.5	1.5	0.500000
2	0.25	1.75	0.250000
3	0.125	1.875	0.125000
4	0.0625	1.9375	0.062500
5	0.03125	1.96875	0.031250
8	0.003906	1.996094	0.003906
10	0.000977	1.999023	0.000977
∞	→ 0	2 (exact)	0

Notice the error at step n equals exactly ar^(n+1): the first term not yet included. The partial sum always falls short of S∞ by precisely one more geometric term.

06 — Zeno's Paradox: Why Infinite Steps Can Take Finite Time

Around 450 BC, the Greek philosopher Zeno of Elea posed a paradox: to cross a room, you must first cross half of it. Then half of what remains. Then half again. Since there are infinitely many steps, surely you can never reach the other side?

Zeno's Paradox, Infinite Halvings on a Number Line

Each step covers half the remaining distance. The geometric series ½ + ¼ + ⅛ + ⋯ = 1/(1 − ½) − 1 = 1. Zeno's paradox is resolved: infinitely many steps of shrinking size can have a finite total.

Mathematically: Σ_n=1^∞ (1/2)ⁿ = (1/2)/(1 − 1/2) = 1. The series converges to 1 because, while there are infinitely many terms, they get small fast enough to sum to a finite value.

07 — Real-World Application: Present Value & Finance

Geometric series are the backbone of financial mathematics. If you receive £P at the end of each year forever (a perpetuity), and money can earn an annual interest rate of i, the present value of all those payments is:

Perpetuity — Present Value Formula (derived from geometric series)

PV = P/(1+i) + P/(1+i)² + P/(1+i)³ + ⋯ = P/i

This is a geometric series with a = P/(1+i) and r = 1/(1+i). Since i > 0 means |r| < 1, the series converges.

Example: A pension paying £5,000/year forever at 5% annual interest has PV = 5000/0.05 = £100,000 today.

The same logic underlies mortgage calculations, bond pricing, and GDP multiplier effects in economics: all involve summing a geometric series of discounted future cash flows.

08 — Ten Worked Examples

EX 01 Infinite sum — basic application of S = a/(1−r)

Find the sum of 1 + 1/2 + 1/4 + 1/8 + ⋯

Identify: a = 1, r = 1/2. Check |r| = 1/2 < 1 ✓ — series converges.

Apply: S = a / (1 − r) = 1 / (1 − 1/2) = 1 / (1/2)

S = 2

EX 02 Identifying a and r from the series

Find the sum: 6 − 2 + 2/3 − 2/9 + ⋯

Identify: a = 6. Ratio: r = −2/6 = −1/3. Check: |r| = 1/3 < 1 ✓

Verify second term: 6 × (−1/3) = −2 ✓, third term: −2 × (−1/3) = 2/3 ✓

S = 6 / (1 − (−1/3)) = 6 / (4/3) = 6 × 3/4

S = 18/4 = 9/2 = 4.5

EX 03 Partial sum — sum of first 6 terms

Find S₅ (the sum of the first 6 terms, n = 0 to 5) of 3 + 6 + 12 + 24 + ⋯

a = 3, r = 2. (Note: |r| = 2 > 1, so the infinite sum diverges — but partial sums are still finite.)

S₅ = 3 · (1 − 2⁶) / (1 − 2) = 3 · (1 − 64) / (−1)

= 3 · (−63) / (−1) = 3 × 63

S₅ = 189

EX 04 Series written in sigma notation

Find the sum: Σ_n=0^∞ 5 · (2/3)ⁿ

Read off: a = 5 (the coefficient when n = 0), r = 2/3. |r| < 1 ✓

S = 5 / (1 − 2/3) = 5 / (1/3)

S = 15

EX 05 Series starting at n = 1 instead of n = 0

Find Σ_n=1^∞ 4 · (1/3)ⁿ

The full series from n = 0 would be: S_full = 4/(1 − 1/3) = 4/(2/3) = 6

Subtract the n = 0 term: 4 · (1/3)⁰ = 4.

S = 6 − 4 = 2

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Alternatively: first term is 4·(1/3)¹ = 4/3, same r = 1/3. S = (4/3)/(1 − 1/3) = (4/3)/(2/3) = 2 ✓

S = 2

EX 06 Recurring decimal as a geometric series

Express 0.̄7̄2̄ = 0.727272… as a fraction using a geometric series.

Write as a series: 0.72 + 0.0072 + 0.000072 + ⋯

a = 0.72 = 72/100, r = 1/100. |r| < 1 ✓

S = (72/100) / (1 − 1/100) = (72/100) / (99/100) = 72/99

Simplify: GCD(72, 99) = 9. 72/99 = 8/11

0.̄7̄2̄ = 8/11 — verify: 8 ÷ 11 = 0.727272… ✓

EX 07 Finding r given the sum and first term

An infinite geometric series has first term a = 8 and sum S = 24. Find r.

Use S = a/(1 − r): 24 = 8 / (1 − r)

Rearrange: 1 − r = 8/24 = 1/3

r = 1 − 1/3 = 2/3

Check |r| = 2/3 < 1 ✓ (consistent with convergence).

r = 2/3

EX 08 Divergence — recognising when the sum doesn't exist

Does Σ_n=0^∞ 3 · (−1.2)ⁿ converge? If so, find its sum.

Identify r = −1.2. Compute |r| = 1.2.

Since |r| = 1.2 > 1, the series diverges.

The series diverges — no finite sum exists. Terms grow in magnitude: 3, −3.6, 4.32, −5.184, …

EX 09 Finance — present value of a perpetuity

An endowment pays £2,000 at the end of each year forever. If the discount rate is 4% per year, what is the present value?

PV = 2000/(1.04) + 2000/(1.04)² + ⋯ — a geometric series with a = 2000/1.04 and r = 1/1.04.

|r| = 1/1.04 < 1 ✓. Apply S = a/(1 − r):
PV = (2000/1.04) / (1 − 1/1.04) = (2000/1.04) / (0.04/1.04)

= 2000 / 0.04

PV = £50,000 — the perpetuity formula P/i handles this in one step.

EX 10 Bouncing ball — total distance travelled

A ball is dropped from a height of 10 m. After each bounce it rises to 60% of its previous height. Find the total distance the ball travels.

First fall: 10 m (down). Then it bounces up and comes back down — each such cycle is 2 × height.

After the first bounce, heights are: 6, 3.6, 2.16, … — geometric with a = 6, r = 0.6.

Total bounce distance (up + down) = 2 × S = 2 × 6/(1 − 0.6) = 2 × 15 = 30 m.

Add the initial drop: Total = 10 + 30 = 40 m.

Total distance = 40 m — alternatively: 10 × (1 + 0.6)/(1 − 0.6) = 10 × 1.6/0.4 = 40 m (shortcut).

09 — Common Mistakes

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Mistake 1 — Applying S = a/(1−r) when |r| ≥ 1: The formula only works for convergent series. Always check |r| < 1 before computing. Writing S = 1/(1−2) = −1 for a divergent series is a serious error.

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Mistake 2 — Miscounting the starting index: If the sum starts at n = 1 (not n = 0), the first term is ar, not a. Either adjust a accordingly, or compute the n = 0 sum and subtract the n = 0 term.

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Mistake 3 — Confusing the partial sum and infinite sum formulas: Sₙ = a(1 − rⁿ⁺¹)/(1 − r) includes n+1 terms. For the infinite sum with |r| < 1, simply let n → ∞ so rⁿ⁺¹ → 0, leaving S = a/(1 − r).

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Mistake 4 — Wrong sign on r: With alternating series like 1 − 1/3 + 1/9 − ⋯, the ratio is r = −1/3 (negative). Using r = +1/3 gives the wrong answer. Always divide the second term by the first to find r, and preserve the sign.

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Next step: Geometric series are the gateway to Power Series, where we replace the constant ratio r with a variable x. Every power series is essentially a generalised geometric series. See also Series Convergence for the broader framework of convergence tests.

🔗 Related Topics — Series & Sequences

Sequences Series Convergence p-Series Test Comparison Tests Ratio & Root Tests Alternating Series Power Series Taylor Series Maclaurin Series Limits at Infinity All Series Topics

Geometric Series
The Only Series with a Closed-Form Sum

01 — What Is a Geometric Series?

02 — The Partial Sum Formula

Deriving the Formula

03 — Infinite Sum and Convergence

Proof of the Infinite Sum Formula

04 — Interactive Convergence Visualiser

05 — Partial Sums Table: Watching Convergence in Action

06 — Zeno's Paradox: Why Infinite Steps Can Take Finite Time

07 — Real-World Application: Present Value & Finance

08 — Ten Worked Examples

09 — Common Mistakes

🔗 Related Topics — Series & Sequences

10 — Quiz: Geometric Series (10 Questions)