TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ the series diverges

Statement

Let $r \in \mathbb{R}$ and let $r^k$ be the integer power (Integer powers $a^m$ ), so that $r^0 = 1$ for every $r$ , including $r = 0$ .

If $|r| < 1$ then the series $\sum r^k$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) and $\sum_{k=0}^{\infty} r^{k} \;=\; \frac{1}{1-r} .$
If $|r| \ge 1$ then $\sum r^k$ diverges.

The series starts at $k = 0$ and its first term is $r^0 = 1$ ; in particular $\sum_{k=0}^{\infty} 2^{-k} = 2$ , while the series starting at $k = 1$ sums to $1$ . Which starting index is meant has to be said, and it is said here.

Facts & Assumptions

Given: A real number $r$ , the integer powers $r^k$ (Integer powers $a^m$ ), and the partial sums $s_n = \sum_{k<n} r^k$ of $\sum r^k$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

Factorisation of a difference of powers: for $a, b \in \mathbb{R}$ and natural $n \ge 1$ , $b^n - a^n = (b-a)\sum_{k=0}^{n-1} a^k b^{\,n-1-k}$ (Factorisation of $b^n - a^n$ , and the resulting Lipschitz estimate).

[L2]

For $|r| < 1$ the sequence $(r^k)$ is null, that is $r^k \to 0$ (For $|r| < 1$ the sequence $r^k$ is null, and for $|r| > 1$ the sequence $|r|^k$ diverges to $+\infty$ ).

[L3]

Algebra of limits: sums, differences and quotients of convergent sequences converge to the corresponding combination, the quotient rule requiring a nonzero limit and nonzero denominators (Algebra of limits: sums, scalar multiples, products and quotients, Limits and Cauchy sequences of reals).

[L4]

Absolute value: $|xy| = |x|\,|y|$ , $|x| \ge 0$ , and $|x| = 0$ exactly when $x = 0$ ; also $|1| = 1$ , since $1 > 0$ (Basic properties of the absolute value).

[L5]

Powers and order: $a^0 = 1$ for every $a$ ; if $a \ge 1$ and $n \ge 1$ then $a^n \ge a \ge 1$ ; and $1^n = 1$ for every $n$ (Monotonicity of $x \mapsto x^n$ and of $n \mapsto a^n$ , Integer powers $a^m$ ).

[L6]

The principle of induction (The principle of mathematical induction).

[L7]

If a series converges then its terms tend to $0$ (If a series converges then its terms tend to $0$ ).

[L8]

Notation of Finite sums and finite products, by recursion: $\sum_{k=0}^{n-1} x_k$ is $\sum_{k<n} x_k$ , and the empty sum $\sum_{k<0} x_k$ is $0$ .

Proof

technique · cases

1.1

Assume $|r| < 1$ .

assume-case lt

1.2

Assume instead $|r| \ge 1$ .

assume-case ge

1.3

For every natural $n \ge 1$ , applying [L1] with $b = 1$ and $a = r$ gives $1 - r^n = (1-r)\sum_{k=0}^{n-1} r^k \cdot 1^{\,n-1-k} = (1-r)\,s_n$ , using $1^m = 1$ and the notation of [L8].

L1L5L8

1.4

At $n = 0$ the identity $1 - r^n = (1-r)s_n$ also holds, both sides being $0$ because $r^0 = 1$ and $s_0$ is the empty sum.

L5L8

2.1

In the case $|r| < 1$ we have $r \ne 1$ , since $|1| = 1$ and $|r| < 1$ ; hence $1 - r \ne 0$ .

step 1.1L4algebra

2.2

In the case $|r| \ge 1$ , an induction gives $|r^k| = |r|^k$ for every $k \in \mathbb{N}$ : at $k = 0$ both sides are $1$ , and if $|r^k| = |r|^k$ then $|r^{k+1}| = |r^k \cdot r| = |r^k|\,|r| = |r|^k |r| = |r|^{k+1}$ .

step 1.2L4L5L6

2.3

In the case $|r| \ge 1$ we get $|r|^k \ge 1$ for every $k \in \mathbb{N}$ : at $k = 0$ this reads $1 \ge 1$ , and for $k \ge 1$ it is the comparison $|r|^k \ge |r| \ge 1$ .

step 1.2L5

3.1

In the case $|r| < 1$ , dividing by $1 - r \ne 0$ gives $s_n = (1 - r^n)/(1-r)$ for every $n \in \mathbb{N}$ .

step 2.1step 1.3step 1.4algebra

3.2

In the case $|r| \ge 1$ , combining the two previous steps gives $|r^k - 0| = |r^k| = |r|^k \ge 1$ for every $k \in \mathbb{N}$ .

step 2.2step 2.3

4.1

In the case $|r| < 1$ the sequence $(r^n)$ is null, so $1 - r^n \to 1$ and therefore $s_n \to 1/(1-r)$ , the denominator being the nonzero constant $1-r$ ; hence $\sum r^k$ converges with sum $1/(1-r)$ , which is claim 1.

step 1.1step 3.1step 2.1L2L3

4.2

In the case $|r| \ge 1$ the sequence $(r^k)$ does not converge to $0$ , since the rational tolerance $\varepsilon = 1$ admits no index $K$ with $|r^k - 0| < 1$ for all $k \ge K$ ; so by the term test $\sum r^k$ diverges, which is claim 2.

step 3.2L7

5.1

The two cases $|r| < 1$ and $|r| \ge 1$ exhaust the possibilities, since the order on $\mathbb{R}$ is total, so claims 1 and 2 together cover every real $r$ .

step 4.1step 4.2cases-exhaustive∎

Remarks

The divergence half needs no separate treatment of $r = 1$ and $r = -1$ . Both are covered by $|r| \ge 1$ , and the single reason is the same in every case: the terms have absolute value at least $1$ , so they cannot tend to $0$ . For $r = 1$ the partial sums are $s_n = n$ and run to $+\infty$ ; for $r = -1$ they oscillate between $0$ and $1$ . The theorem says only that neither converges, which is all that "diverges" means here (Series, partial sums, convergence and the sum, divergence, and the tail series).
Why the identity is proved at $n = 0$ separately. Factorisation of $b^n - a^n$ , and the resulting Lipschitz estimate requires $n \ge 1$ , since its right-hand side is a sum over $k < n$ of a term involving $b^{\,n-1-k}$ , and $n-1$ is not a natural number at $n = 0$ . The identity is still true at $n = 0$ , but by inspection of two empty objects rather than by that lemma, and step 1.4 says so rather than letting the reader assume the citation covers it.

Depends on

Used by

…and 16 more results.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 93 results over 26 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Geometric series (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

For ∣r∣<1|r| < 1∣r∣<1, ∑k≥0rk=1/(1−r)\sum_{k \ge 0} r^k = 1/(1-r)∑k≥0​rk=1/(1−r), and for ∣r∣≥1|r| \ge 1∣r∣≥1 the series diverges