The geometric series adds up

Adam Zettel · September 22, 2026 · written with an agent

S_{n} = a \frac{1 - r ^{n}}{1 - r} \to \frac{a}{1 - r}

$1 + \frac{1}{2} + \frac{1}{4} + \dots$ never reaches 2, and the partial sums say exactly how far short: $S_{n} = 2 - 2^{1 - n}$ . The general fact is $S_{n} = a \frac{1 - r ^{n}}{1 - r}$ , and when $∣ r ∣ < 1$ the $r^{n}$ goes to zero and the whole series is $a / (1 - r)$ . The program adds the terms one by one and checks the formula at every $n$ .

Open in the workbench Download .vivOutputs below are from the site build’s run; Run all runs it again in your browser.

Look inside this program

Atlas
The program’s map: every function and which calls which.
There’s no recorded run of this program yet.

The trick that gives the formula: multiply $S_{n} = a + a r + \dots + a r^{n - 1}$ by $r$ and subtract. Every term cancels except the first of $S_{n}$ and the last of $r S_{n}$ :

S_{n} - r S_{n} = a - a r^{n} ⟹ S_{n} = a \frac{1 - r ^{n}}{1 - r} .

The figure draws the running sums for $r = 1/2$ and $r = 0.9$ against their limits 2 and 10. The $r = 0.9$ series takes 44 terms to get within 1% of its limit; the $r = 1/2$ series takes 7. The tests check the formula at every $n$ up to 50, how close fifty terms of the $r = 0.9$ series get to 10, and one more series with $a = 3$ , $r = 0.2$ .

Printed

1.99805

9.94846

Running sums of a geometric series

r = 1/2, limit 2r = 0.9, limit 10

Tests 5 of 5 pass

the running sum matches a (1 - r^n) / (1 - r) at every n, for both ratios (passes)
ten terms of 1 + 1/2 + 1/4 + ... come to 2 - 2^-9 = 1.998046875, exactly (passes)
the r = 0.9 series takes 44 terms to get within 1% of its limit, and the r = 1/2 series takes 7 (passes)
fifty terms of the r = 0.9 series are within 0.6% of the limit 10 (passes)
a = 3, r = 0.2: five terms give 3.7488 and the limit is 3.75 (passes)