Infinite Geometric Series

An infinite geometric series \(a + a\cdot r + a\cdot r^2 + \dots\) keeps adding terms forever, yet under the right condition on the common ratio r it still settles on a single finite total.

By the end you'll be able to tell whether an infinite geometric series converges or diverges, and compute its sum \(S=a/(1-r)\) when it does.

Predict: will \(2 + 2r + 2r^2 + \dots\) settle on a finite total for r = 0.5? What about r = 2? Set r on the slider and check.

Each bar stacks the running total (partial sum) after n terms: the dim base is everything summed before, the bright top segment is the newest term just added. Watch the bars as you drag r — below r = 1 the added slivers shrink and the bars level off at the dashed limit line \(S=a/(1-r)\); at r ≥ 1 the added pieces stay the same size or grow, and the bars just keep climbing (some run off the top of the chart — that's divergence).

a = 2, r = 0.50 → converges: S = a/(1−r) = 2/0.50 = 4.00

Partial sums of 2 + 2r + 2r² + … — 8 bars, dashed line at the limit S = a/(1−r)
already summed newest term (converging) newest term (diverging)

Add up infinitely many numbers and you'd expect an infinite total — but a geometric series can defy that, landing on one finite sum whenever its terms shrink fast enough.

Intuitive

Picture the bars from the diagram above: each new term is a slice added on top of the running total. If every slice is smaller than the last (|r| < 1), the slices pile up but get thinner and thinner, so the stack's height keeps closing in on a ceiling — that ceiling is the sum. If the slices stay the same size or grow (|r| ≥ 1), there's no ceiling to close in on, so the stack just keeps climbing.

Formal

Start from the finite-sum formula for the first n terms, the partial sum S_n: \(S_n = a\cdot\dfrac{1-r^n}{1-r}\). The only piece that depends on n is \(r^n\). When \(|r|<1\), raising r to higher powers drives \(r^n\) toward 0, so \(S_n\) settles down to \(S = a/(1-r)\) — the series converges. When \(|r|\ge 1\), \(r^n\) does not shrink to 0 (it stays the same size or grows), so \(S_n\) never settles — the series diverges and has no finite sum.

Applied

Repeating decimals are infinite geometric series in disguise: \(0.333\ldots = 0.3 + 0.03 + 0.003 + \dots\) has a = 0.3 and r = 0.1, so \(|r|<1\) and it converges to \(0.3/(1-0.1) = 0.3/0.9 = 1/3\) — confirming what you already know about \(1/3\) as a decimal.

Worked example

Evaluate \(8 + 4 + 2 + 1 + \tfrac{1}{2} + \dots\)  Step 1 — identify a and r: \(a=8\), \(r = 4/8 = 1/2\). Step 2 — check convergence: \(|r|=1/2 < 1\), so it converges and the formula applies. Step 3 — apply \(S=a/(1-r)\): \(S = 8/(1-\tfrac12) = 8/(\tfrac12) = \) 16. The total is finite even though the terms go on forever — each new term adds at most half of what remains.

Your turn

Same idea, different numbers: \(3 + 0.6 + 0.12 + \dots\) Step 1: \(a=3\), \(r = 0.6/3 = 0.2\). Step 2: \(|r|=0.2 < 1\), so it converges. Step 3: apply \(S=a/(1-r) = 3/(1-0.2) = 3/0.8 = \) ____

Reveal the answer

\(S = 3/0.8 = \) 3.75. Set r near 0.2 on the slider above (a is fixed at 2 there, so the exact numbers differ, but the shrinking-slice pattern is the same) and watch the bars close in on a small ceiling, just like this one.

More info — why the finite-sum formula turns into \(a/(1-r)\)

The finite sum of the first n terms of a geometric sequence is \(S_n = a\cdot(1-r^n)/(1-r)\). Letting n grow forever, the only thing that changes is \(r^n\): when \(|r|<1\) it shrinks toward 0, so the formula's numerator \((1-r^n)\) approaches 1 and \(S_n\) approaches \(a/(1-r)\) — that's the whole derivation. When \(|r|\ge 1\), \(r^n\) doesn't shrink, so there's nothing for \(S_n\) to approach. See the Math is Fun link in Dive deeper for the full step-by-step derivation.

Check your understanding

Question 1 of 4

Evaluate \(9 + 3 + 1 + \tfrac{1}{3} + \dots\)

Question 2 of 4

Predict: for the series \(6 + 6r + 6r^2 + \dots\), does it converge when r = 0.5? What about r = 1.5?

Question 3 of 4

What happens to \(a + a\cdot r + a\cdot r^2 + \dots\) (with \(a \ne 0\)) when r = 1?

Question 4 of 4

A geometric sequence has \(a_1 = 4\) and common ratio \(r = 1/4\) (so \(a_2=1\), \(a_3=0.25\), …). What is the sum of ALL infinitely many terms?

Recap

  • An infinite geometric series is \(a + a\cdot r + a\cdot r^2 + \dots\), built from a geometric sequence's terms.
  • It converges to \(S = a/(1-r)\) when \(|r| < 1\) — the terms shrink toward 0, so the partial sums settle down.
  • It diverges (no finite sum) when \(|r| \ge 1\) — the terms don't shrink, so the partial sums never settle.
  • Always check \(|r| < 1\) before applying the formula — it never applies at r = 1.

Dive deeper

Sources

  • Infinite Geometric Series