Notebooks · posts/the-geometric-series-adds-up

The geometric series adds up

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

$1 + \frac{1}{2} + \frac{1}{4} + \dots$ 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 $n$ .

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

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 $a = 3$ , $r = 0.2$ .

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[1]lines 23–90

N = 50

#| $S_1, \ldots, S_n$ by adding the terms one at a time.
def partial_sums(a, r, n) {
    "S_1 .. S_n by adding the terms one at a time."
    s = zeros(n)
    total = 0.0
    term = a
    for i to n {
        total += term
        s[i] = total
        term = term * r
    }
    return s
}
#| $S_n = a\,\frac{1 - r^n}{1 - r}$, from $S_n - rS_n = a - ar^n$.
def closed_form(a, r, n) {
    "a (1 - r^n) / (1 - r)."
    return a * (1.0 - pow(r, n)) / (1.0 - r)
}

half = partial_sums(1.0, 0.5, N)
nine = partial_sums(1.0, 0.9, N)
show half[9]
show nine[N - 1]

ns = linspace(1.0, N, N)
figure("Running sums of a geometric series", "terms added", "sum")
plot(ns, half, "r = 1/2, limit 2")
plot(ns, nine, "r = 0.9, limit 10")

test "the running sum matches a (1 - r^n) / (1 - r) at every n, for both ratios" {
    for i to N {
        assert_near(half[i], closed_form(1.0, 0.5, i + 1), 1.0e-12)
        assert_near(nine[i], closed_form(1.0, 0.9, i + 1), 1.0e-9)
    }
}
test "ten terms of 1 + 1/2 + 1/4 + ... come to 2 - 2^-9 = 1.998046875, exactly" {
    # the third term is 1/4
    assert_near(half[2] - half[1], 0.25, 0.0)
    assert_near(half[9], 1.998046875, 1.0e-14)
}
test "the r = 0.9 series takes 44 terms to get within 1% of its limit, and the r = 1/2 series takes 7" {
    # the number of terms before the running sum is first within 1% of the limit
    nine_terms = 0
    half_terms = 0
    for i to N {
        if nine_terms == 0 and abs(nine[i] - 10.0) < 0.01 * 10.0 {
            nine_terms = i + 1
        }
        if half_terms == 0 and abs(half[i] - 2.0) < 0.01 * 2.0 {
            half_terms = i + 1
        }
    }
    assert_near(nine_terms, 44, 0.0)
    assert_near(half_terms, 7, 0.0)
}
test "fifty terms of the r = 0.9 series are within 0.6% of the limit 10" {
    assert_lt(abs(nine[N - 1] - 10.0) / 10.0, 0.006)
}
test "a = 3, r = 0.2: five terms give 3.7488 and the limit is 3.75" {
    s = partial_sums(3.0, 0.2, 5)
    assert_near(s[4], 3.7488, 1.0e-12)
    assert_near(3.0 / (1.0 - 0.2), 3.75, 1.0e-12)
}

# SPDX-License-Identifier: MIT OR Apache-2.0
# SPDX-FileCopyrightText: 2026 The Viv authors

N = 50

#| $S_1, \ldots, S_n$ by adding the terms one at a time.
def partial_sums(a, r, n) {
    "S_1 .. S_n by adding the terms one at a time."
    s = zeros(n)
    total = 0.0
    term = a
    for i to n {
        total += term
        s[i] = total
        term = term * r
    }
    return s
}
#| $S_n = a\,\frac{1 - r^n}{1 - r}$, from $S_n - rS_n = a - ar^n$.
def closed_form(a, r, n) {
    "a (1 - r^n) / (1 - r)."
    return a * (1.0 - pow(r, n)) / (1.0 - r)
}

half = partial_sums(1.0, 0.5, N)
nine = partial_sums(1.0, 0.9, N)
show half[9]
show nine[N - 1]

ns = linspace(1.0, N, N)
figure("Running sums of a geometric series", "terms added", "sum")
plot(ns, half, "r = 1/2, limit 2")
plot(ns, nine, "r = 0.9, limit 10")

test "the running sum matches a (1 - r^n) / (1 - r) at every n, for both ratios" {
    for i to N {
        assert_near(half[i], closed_form(1.0, 0.5, i + 1), 1.0e-12)
        assert_near(nine[i], closed_form(1.0, 0.9, i + 1), 1.0e-9)
    }
}
test "ten terms of 1 + 1/2 + 1/4 + ... come to 2 - 2^-9 = 1.998046875, exactly" {
    # the third term is 1/4
    assert_near(half[2] - half[1], 0.25, 0.0)
    assert_near(half[9], 1.998046875, 1.0e-14)
}
test "the r = 0.9 series takes 44 terms to get within 1% of its limit, and the r = 1/2 series takes 7" {
    # the number of terms before the running sum is first within 1% of the limit
    nine_terms = 0
    half_terms = 0
    for i to N {
        if nine_terms == 0 and abs(nine[i] - 10.0) < 0.01 * 10.0 {
            nine_terms = i + 1
        }
        if half_terms == 0 and abs(half[i] - 2.0) < 0.01 * 2.0 {
            half_terms = i + 1
        }
    }
    assert_near(nine_terms, 44, 0.0)
    assert_near(half_terms, 7, 0.0)
}
test "fifty terms of the r = 0.9 series are within 0.6% of the limit 10" {
    assert_lt(abs(nine[N - 1] - 10.0) / 10.0, 0.006)
}
test "a = 3, r = 0.2: five terms give 3.7488 and the limit is 3.75" {
    s = partial_sums(3.0, 0.2, 5)
    assert_near(s[4], 3.7488, 1.0e-12)
    assert_near(3.0 / (1.0 - 0.2), 3.75, 1.0e-12)
}

# SPDX-License-Identifier: MIT OR Apache-2.0
# SPDX-FileCopyrightText: 2026 The Viv authors