What Is a Geometric Series?
A geometric series adds terms that each differ from the last by a constant multiplier, the common ratio r:
a + ar + ar² + ar³ + ⋯
It's the most completely understood infinite series in mathematics: both its partial sums and its infinite sum have exact closed formulas. The N-term partial sum telescopes to SN = a(1 − rᴺ)/(1 − r), and when |r| < 1 the rᴺ term dies away as N grows, leaving the famous infinite sum:
S = a / (1 − r), provided |r| < 1
When |r| ≥ 1 the terms refuse to shrink, and no amount of adding settles anywhere, the series diverges. The dividing line is razor-sharp: r = 0.999 converges (slowly, to 1000a), r = 1 diverges.
How to Use This Calculator
Enter the first term, the ratio, and how many terms to sum. The calculator applies the partial-sum formula, runs the |r| < 1 convergence test, and reports the infinite sum when it exists, with the running partial sums plotted so you can watch them glide toward the limit (or flee it, for the diverging chip). The 0.999… chip settles a famous internet argument: 0.9 + 0.09 + 0.009 + ⋯ is a geometric series with a = 0.9, r = 0.1, and its sum is exactly 0.9/(1 − 0.1) = 1.
Worked Example
Sum 1 + ½ + ¼ + ⅛ + ⋯ (a = 1, r = ½):
S = 1 / (1 − ½) = 2
The partial sums march 1, 1.5, 1.75, 1.875, … always halving the remaining gap to 2, Zeno's dichotomy paradox resolved in one line. Walking half the distance, then half of what remains, forever, covers precisely the whole distance: infinitely many steps, finite total, because the steps shrink geometrically.
Where the Formula Comes From
The derivation is a two-line gem worth knowing. Write SN = a + ar + ⋯ + ar^(N−1), multiply both sides by r, and subtract:
SN − rSN = a − arᴺ → SN(1 − r) = a(1 − rᴺ)
Every interior term cancels in the subtraction, the telescoping that makes geometric series exactly solvable while almost no other series is. Let N → ∞ with |r| < 1 and rᴺ → 0, delivering S = a/(1 − r).
Common Mistakes to Avoid
- Using a/(1−r) when |r| ≥ 1. The formula's derivation needs rᴺ → 0. Plugging r = 2 into it produces −a, a nonsense "sum" for a series racing to infinity.
- Misidentifying a. The formula's a is the first term actually being summed. If the series starts at n = 1 with terms 3(½)ⁿ, the first term is 3/2, not 3.
- Confusing the ratio with the difference. Geometric series multiply by r each step; arithmetic sequences add a constant. An arithmetic "series" 1 + 2 + 3 + ⋯ diverges always, no closed infinite sum exists to misuse.
- Expecting fast convergence when r is near 1. r = 0.99 converges, but needs hundreds of terms per digit, the partial-sum plot makes the crawl visible.
Real-World Applications
The a/(1−r) formula is quietly one of the most economically consequential lines of mathematics. A perpetuity, a fixed payment forever, discounted at rate i, is a geometric series with r = 1/(1+i), and its sum (payment/i) prices annuities, preferred stocks, and underlies every dividend-discount valuation model. The money multiplier in banking (each deposited dollar partially re-lent, re-deposited, re-lent…) is a geometric series; so is the total economic impact of a stimulus with marginal spending rate r, summing to 1/(1−r) times the initial injection.
Physics and engineering use it wherever an effect echoes with constant attenuation: a bouncing ball whose each bounce reaches fraction r of the last covers total distance governed by the formula (and, remarkably, finishes bouncing in finite time); repeating decimals convert to fractions through it; signal reflections in transmission lines and multiple scattering of light between surfaces sum geometrically; and drug accumulation under repeated doses with fractional elimination reaches exactly the steady state a/(1−r) predicts, the pharmacokinetic plateau every dosing schedule is designed around.
Frequently Asked Questions
Why exactly does |r| < 1 decide convergence?
Because rᴺ shrinks to zero precisely when |r| < 1, the only leftover in the partial-sum formula. At |r| = 1 the terms never shrink (r = 1 adds a forever; r = −1 oscillates), and beyond it they grow. It's also the textbook example behind the ratio test whose L is literally |r| here.
Is 0.999… really equal to 1?
Yes, it's the geometric series 9/10 + 9/100 + ⋯ with sum (9/10)/(1 − 1/10) = 1 exactly. Not approximately, not "in the limit as a hedge": the infinite decimal names the series' sum, and the sum is 1.
What about negative ratios?
Alternating geometric series (the r = −½ chip) converge under the same |r| < 1 rule, with partial sums hopping alternately above and below the limit a/(1−r), for a = 1, r = −½, the sum is 2/3, approached in a shrinking zigzag visible in the plot.
How does this connect to power series?
Substituting the variable x for r turns the sum formula into 1/(1−x) = 1 + x + x² + ⋯, the fundamental power series from which many Maclaurin expansions are built by substitution, differentiation, and integration. The geometric series is the seed of the whole power-series orchard.
How many terms until the partial sum is "close enough"?
The leftover after N terms is exactly arᴺ/(1−r), so solve |r|ᴺ ≤ tolerance: N ≥ log(tol)/log|r|. For r = ½, ten terms land within a thousandth of the sum; for r = 0.9, the same accuracy needs 66 terms, geometric decay's speed is set entirely by how far r sits from 1.
What if my series starts at n = 0 versus n = 1?
Only the value of "a" changes: Σ from n = 0 of arⁿ has first term a, while starting at n = 1 makes it ar. The formula S = (first term)/(1 − r) is index-proof if you always feed it the first term actually summed, the safest habit for exam setups.