Drop a bouncing ball and each bounce rises a fixed fraction of the last; take a repeated dose of medicine and each one decays by the same factor before the next; mortgages and savings plans grow the same way. All of these are one kind of sum — every term a fixed multiple of the one before — and, remarkably, it is the one infinite sum we can total exactly.

Here is an uncomfortable secret about infinite series: for almost all of them, we can prove they converge without ever being able to say what they converge to. The exact value of \sum 1/n^3 has no known closed form; even \sum 1/n^2 = \pi^2/6 took Euler and a stroke of genius. Against this backdrop stands one glorious exception — a whole family of series we can sum exactly, with a one-line formula, from a proof short enough to fit on a napkin.

That family is the geometric series: an infinite sum whose terms form a sequence in which each term is a fixed ratio r times the last,

with first term a \ne 0 and common ratio r. Whether this converges turns entirely on the size of r: if |r| < 1 each term is a fixed fraction of the last and the total settles; if |r| \ge 1 the terms refuse to shrink and the sum runs away. And when it does converge, the value is a single clean formula, a/(1-r).

That would already make it worth knowing, but its real importance runs deeper: the geometric series is the yardstick of convergence. The great tests you will meet next — the ratio test, the root test, comparison — all work by quietly asking "does this series shrink at least as fast as a geometric one?" Master this page and you hold the reference object that the entire theory of series is measured against. (If you met geometric sums at school, that was the algebraic version; here we do the analysis honestly — partial sums, limits, and a genuine if-and-only-if.)

The centerpiece: closed form, then the limit

We derive everything from scratch, and the strategy matters as much as the result. A series is defined as the limit of its partial sums — so the honest plan is: first find an exact formula for the partial sum s_N (a finite object, no infinity anywhere), and only then let N \to \infty. The formula falls out of a single trick — multiply by r and subtract — which makes the middle of the sum telescope: collapse like a folding telescope, leaving only the two ends.

Step 1 — Write the partial sum. Add the first N+1 terms (indices 0 through N):

Step 2 — Multiply the whole thing by r. Every term's power steps up by one:

Step 3 — Subtract. Line up s_N - r\,s_N: every middle term ar, ar^2, \dots, ar^N appears in both lists and cancels, leaving only the first term of s_N and the last of r s_N:

Step 4 — Factor and solve for s_N. The left side is s_N(1 - r); the right is a(1 - r^{N+1}). Provided r \ne 1 we may divide:

That is the closed form for the partial sum — exact, valid for every r \ne 1, convergent or not, with no infinity in sight yet. Everything that follows is just reading off what this formula does as N grows.

Step 5 — Take N \to \infty, case by case. The only place N appears is the single term r^{N+1}, so the entire fate of the series hangs on the limit of the powers of r:

• |r| < 1: the powers collapse, r^{N+1} \to 0 (multiply any number by something of size below 1 often enough and it is crushed toward zero). The partial sums converge.
• |r| > 1: the powers explode, |r|^{N+1} \to \infty, and s_N blows up with them. Divergent.
• r = 1: the formula was never valid here (we divided by 1-r), so go back to the definition: s_N = a + a + \cdots + a = a(N+1), which marches off to \pm\infty. Divergent.
• r = -1: the partial sums are a,\ 0,\ a,\ 0,\ \ldots — bouncing forever between two values, never settling on either. A sequence with two distinct cluster points has no limit. Divergent.

Step 6 — Read off the sum when |r| < 1. The r^{N+1} term vanishes in the limit, and the partial sums converge:

Notice what we have actually proved: not just a formula, but a complete classification. The series converges if and only if |r| < 1 — sufficiency from the vanishing power, necessity because in every other case we exhibited the divergence explicitly. (For |r| \ge 1 the terms ar^n don't even tend to 0, which the divergence test will tell you dooms any series instantly.) The dividing line is exactly |r| = 1, and the boundary itself belongs to the losing side.

Slide r across the convergence boundary

The whole case analysis above is visible in one picture. Here are the partial sums s_N of \sum a r^n with a = 1, plotted as dots against N. Move the ratio r and watch the four regimes of Step 5 play out:

• For 0 < r < 1 the dots climb and level off at the dashed line \tfrac{1}{1-r} — each new term closes a fixed fraction of the remaining gap, so the staircase flattens into the limit.
• For -1 < r < 0 they spiral in, overshooting and undershooting the limit alternately from both sides — convergence by damped oscillation.
• Push r just past 1 (or below -1) and the picture snaps: the limit line vanishes and the dots run off the frame.
• Park r exactly on 1 or -1: a rising straight line, or a permanent two-point bounce. Neither settles — the boundary always loses.

Notice also how the limit \tfrac{1}{1-r} itself behaves: as r \to 1^- the dashed line shoots upward — the sum still exists, but it takes ever longer to get there. Slow shrinking means slow convergence.

Worked examples: the formula in action

Example 1 — a straight computation. Sum \displaystyle \sum_{n=0}^{\infty} 5\left(\tfrac{2}{3}\right)^{n}. Identify the pieces: a = 5, r = \tfrac{2}{3}, and |r| < 1 so we may sum:

Sanity-check against the partial sums: s_0 = 5, s_1 \approx 8.33, s_2 \approx 10.56, s_3 \approx 12.04, s_{10} \approx 14.85 — creeping up on 15, the remaining gap shrinking by a factor of \tfrac{2}{3} at every step. Indeed the leftover after N terms is exactly 15 - s_N = 15\,(\tfrac23)^{N+1} — the tail of a geometric series is itself geometric.

Example 2 — the classic: a repeating decimal. Convert 0.727272\ldots to a fraction. Slice the decimal into its repeating blocks:

A geometric series with a = \tfrac{72}{100} and r = \tfrac{1}{100}. Apply the formula:

The pattern generalises instantly: a repeating two-digit block d gives d/99, a three-digit block gives d/999, and so on — the string of nines is just 10^k - 1 emerging from 1 - r.

Example 3 — when the first term isn't the obvious one. Sum \displaystyle \sum_{n=1}^{\infty} 3\left(\tfrac{1}{4}\right)^{n}. The index starts at n = 1, so the first term actually present is 3 \cdot \tfrac14 = \tfrac34 — that is a, not 3. With a = \tfrac34, r = \tfrac14:

Cross-check by the other road: the full sum from n = 0 is 3/(1 - \tfrac14) = 4; subtract the missing n = 0 term, 3, and again 4 - 3 = 1. Two routes, one answer — and a habit worth forming: a is always the first term that is actually there.

Example 4 — the bouncing ball. A ball is dropped from 2 m and each bounce carries it back up to \tfrac{3}{5} of its previous height. Total distance travelled? The drop contributes 2; then every bounce height 2(\tfrac35)^n (for n \ge 1) is travelled twice — up and down:

An infinite number of bounces, a finite distance — and, if you carry the same analysis through the physics, a finite time too (the times form another geometric series, with ratio \sqrt{3/5}). The ball genuinely stops.

The yardstick of the whole theory

Why does this one family get a page of its own in an analysis course? Because it is the series everything else is compared to. The ratio test asks whether the ratios of consecutive terms eventually stay below some r < 1 — that is, whether the tail of your series is dominated by a convergent geometric one. The root test does the same with n-th roots, and the comparison tests let you make such dominance arguments by hand. In each case the geometric series supplies the standard of "shrinking fast enough": we can't sum most series exactly, but we can trap them beneath one we can.

And the formula itself has a second life. Read \frac{1}{1-x} = \sum_{n=0}^{\infty} x^{n} not as a fact about one number x but as an identity between a function and a series of powers, valid on the interval |x| < 1 — and you are holding the first power series, the seed from which the representations of e^x, \ln(1+x) and \arctan x all grow. One napkin-length telescope argument, and half of analysis leans on it.

See it explained