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✓ 8 results · all verified · 0 also independently AI-judged

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Power Series and Real-Analytic Functions — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

The geometric series represents $1 / (1 - x)$ for $∣ x ∣ < 1$ and re-expands explicitly about every $c$ with $∣ c ∣ < 1$

Statement

For $∣ x ∣ < 1$ ,

$\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^{n} .$

More generally, if $∣ c ∣ < 1$ , then

$\frac{1}{1 - x} = \sum_{n = 0}^{\infty} \frac{(x - c)^{n}}{(1 - c)^{n + 1}} (∣ x - c ∣ < 1 - c) .$

Facts & Assumptions

Given: A real $c$ with $∣ c ∣ < 1$ .

[L1]

The geometric series sums to $1 / (1 - t)$ for $∣ t ∣ < 1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[L2]

Power-series sums re-expand about interior points, and a nonzero local denominator has a reciprocal series (A power-series sum may be re-expanded about every interior point, with coefficients given by its derivatives there, A convergent real power series with nonzero constant term has a convergent reciprocal power series on a smaller neighbourhood).

Verification

technique · direct

1.1

Apply [L1] with $t = x$ to get the first formula.

2.1

The general results in [L2] show qualitatively that the sum re-expands about $c$ and that the nonzero denominator there has a local reciprocal series. To identify that series and its full convergence interval directly, use $1 - x = (1 - c) (1 - (x - c) / (1 - c))$ and $1 - c > 0$ , and apply [L1] with $t = (x - c) / (1 - c)$ . This gives the second formula precisely when $∣ x - c ∣ < 1 - c$ .

givenL1L2algebra∎

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

The radius-one series with coefficients $1 / (n + 1)^{2}$ , $1 / (n + 1)$ and $1$ realise absolute, conditional and divergent endpoint behaviour

Statement

Each of

$\sum_{n \geq 0} \frac{x^{n}}{ι (n + 1)^{2}}, \sum_{n \geq 0} \frac{x^{n}}{ι (n + 1)}, \sum_{n \geq 0} x^{n}$

has radius $1$ . At $x = \pm 1$ the first converges absolutely; the second converges conditionally at $- 1$ and diverges at $1$ ; the third diverges at both endpoints.

Facts & Assumptions

Given: The three displayed real power series.

[L1]

The relevant coefficient roots tend to $1$ , because $ι (n)^{1 / n} \to 1$ and limits respect products and reciprocals ( $n^{1 / n} \to 1$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Algebra of limits: sums, scalar multiples, products and quotients).

[L2]

Cauchy–Hadamard converts that limit into radius $1$ (Cauchy–Hadamard: the reciprocal radius is ${lim sup}_{k \to \infty} ∣ a_{k + 1} ∣^{1 / (k + 1)}$ , with the zero and infinite cases included).

[L3]

The $p$ -series converges for rational $p > 1$ and diverges for $p \leq 1$ , while the alternating harmonic series converges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ ).

Verification

technique · cases

1.1

By [L1], all three Cauchy–Hadamard limit superiors equal $1$ , so [L2] gives radius $1$ in each case.

L1L2

1.2

For the squared-denominator series, absolute values at either endpoint form the $p$ -series with $p = 2$ , which converges by [L3].

assume-case squaredL3

1.3

For the first-power denominator, $x = 1$ gives the divergent harmonic series, while $x = - 1$ gives a convergent alternating series whose absolute series is harmonic.

assume-case harmonicL3

1.4

For the constant coefficients, at either endpoint the terms have absolute value $1$ and do not tend to zero, so both endpoint series diverge.

assume-case constantgiven

2.1

The coefficient families in steps 1.2--1.4 exhaust the displayed series and give the asserted absolute, conditional, and divergent endpoint behaviours.

step 1.1step 1.2step 1.3step 1.4cases-exhaustive∎

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

The series $\sum_{n \geq 0} ι (n!) x^{n}$ converges only at $x = 0$ and has radius zero

Statement

The real power series $\sum_{n \geq 0} ι (n!) x^{n}$ has radius $0$ : it converges at its centre $x = 0$ and diverges at every $x \neq 0$ .

Facts & Assumptions

Given: Coefficients $a_{n} : = ι (n!)$ .

[L1]

Factorials eventually exceed every fixed geometric progression (For every real $x$ , $x^{k} / k! \to 0$ , The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L2]

Cauchy–Hadamard assigns radius $0$ when the coefficient-root limit superior is $+ \infty$ (Cauchy–Hadamard: the reciprocal radius is ${lim sup}_{k \to \infty} ∣ a_{k + 1} ∣^{1 / (k + 1)}$ , with the zero and infinite cases included).

Verification

technique · direct

1.1

For every real $M > 0$ , [L1] applied to a geometric base larger than $M$ shows that $ι ((k + 1)!)^{1 / (k + 1)} > M$ eventually. Hence the coefficient-root limit superior is $+ \infty$ .

2.1

By [L2], the radius is $0$ . At $x = 0$ only the constant term survives, while for $x \neq 0$ the terms fail to tend to zero, so the stated convergence set follows.

step 1.1L2∎

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

The geometric series converges pointwise but not uniformly on $(- 1, 1)$

Statement

The partial sums of $\sum_{n \geq 0} x^{n}$ converge pointwise to $1 / (1 - x)$ on $(- 1, 1)$ but do not converge uniformly there.

Facts & Assumptions

Given: The geometric-series partial sums on $(- 1, 1)$ .

[L1]

Pointwise convergence follows from the geometric-series theorem (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[L2]

Uniform convergence implies the uniform Cauchy property (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

Verification

technique · direct

1.1

Consecutive partial sums differ by $x^{N}$ , whose supremum over $(- 1, 1)$ is $1$ for every $N$ .

givenalgebra

2.1

Thus the partial sums are not uniformly Cauchy by [L2] and hence not uniformly convergent, though [L1] gives pointwise convergence. This is the counterexample recorded in FALSE: every power series converges uniformly on its entire open interval of convergence.

step 1.1L1L2∎

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

A rational function with nonvanishing denominator is locally represented by geometric-series expansions

Statement

The rational function $r (x) = 1 / (2 - x)$ is real analytic on $R ∖ {2}$ . At every $c \neq 2$ it has the local expansion

$\frac{1}{2 - x} = \sum_{n = 0}^{\infty} \frac{(x - c)^{n}}{(2 - c)^{n + 1}} (∣ x - c ∣ < ∣ 2 - c ∣) .$

Facts & Assumptions

Given: A centre $c \neq 2$ .

[L1]

The geometric series converges for arguments of absolute value below $1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[L2]

A nonvanishing analytic denominator has a local analytic reciprocal (A convergent real power series with nonzero constant term has a convergent reciprocal power series on a smaller neighbourhood, Real-analytic functions are closed under sums, products and compositions, and under quotients where the denominator is nonzero).

Verification

technique · direct

1.1

Factor $2 - x = (2 - c) (1 - (x - c) / (2 - c))$ and apply [L1]. The resulting series is exactly the displayed one and converges when $∣ x - c ∣ < ∣ 2 - c ∣$ .

givenL1algebra

2.1

Since every $c \neq 2$ admits this positive-radius local representation, $r$ is real analytic on its domain, in agreement with [L2].

step 1.1L2∎

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

The alternating harmonic series illustrates Abel's boundary-limit theorem without evaluating its sum

Statement

Let $s$ denote the ordinary sum of the alternating harmonic series

$s : = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{ι (n + 1)} .$

Then, without evaluating $s$ ,

$\lim_{x ↑ 1} \sum_{n = 0}^{\infty} \frac{(- 1)^{n} x^{n}}{ι (n + 1)} = s .$

Facts & Assumptions

Given: The zero-indexed alternating harmonic series.

[L1]

It converges by the alternating-series test (The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L2]

Abel's limit theorem identifies the boundary limit with the ordinary sum of any convergent series (Abel's limit theorem: if a real series converges to $s$ , then its power series tends to $s$ as $x ↑ 1$ ).

Verification

technique · direct

1.1

By [L1], the ordinary sum $s$ exists.

2.1

Apply [L2] to its coefficients $a_{n} = (- 1)^{n} / ι (n + 1)$ to obtain the displayed limit. No closed-form evaluation of $s$ is needed.

step 1.1L2∎

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

Grandi's series is Abel summable to $1 / 2$ but its partial sums do not converge

Statement

Grandi's series $1 - 1 + 1 - 1 + \dots$ is Abel summable to $1 / 2$ but diverges ordinarily.

Facts & Assumptions

Given: Coefficients $a_{n} = (- 1)^{n}$ .

[L1]

For $0 \leq x < 1$ , $\sum_{n \geq 0} (- 1)^{n} x^{n} = 1 / (1 + x)$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[L2]

Abel summability and ordinary convergence are defined through the boundary limit and partial sums, respectively (Abel summability by $\lim_{x ↑ 1} \sum a_{n} x^{n}$ and Cesaro summability by the Cesaro means of the partial sums).

Verification

technique · direct

1.1

By [L1], the Abel transform tends to $1 / 2$ as $x ↑ 1$ .

L1L2

2.1

The partial sums alternate between $1$ and $0$ , so they diverge. This concretely refutes FALSE: Abel summability alone implies ordinary convergence of a series.

givenL2∎

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

$1 - 2 + 3 - 4 + \dots$ is Abel summable to $1 / 4$ but is not Cesaro summable

Statement

The series $\sum_{n \geq 0} (- 1)^{n} ι (n + 1)$ is Abel summable to $1 / 4$ , but its Cesaro means do not converge.

Facts & Assumptions

Given: Coefficients $a_{n} = (- 1)^{n} ι (n + 1)$ .

[L1]

The geometric series for $1 / (1 + x)$ may be differentiated term by term for $∣ x ∣ < 1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Inside its radius a real power series may be differentiated term by term, and the differentiated series has the same radius).

[L2]

Cesaro means average the inclusive partial sums (The Cesaro means $σ_{n} = (x_{0} + \dots + x_{n}) / (n + 1)$ and $(C, 1)$ -summability, Abel summability by $\lim_{x ↑ 1} \sum a_{n} x^{n}$ and Cesaro summability by the Cesaro means of the partial sums).

Verification

technique · direct

1.1

Differentiating $\sum_{n \geq 0} (- 1)^{n} x^{n} = 1 / (1 + x)$ and combining with the original series gives $\sum_{n \geq 0} (- 1)^{n} ι (n + 1) x^{n} = 1 / (1 + x)^{2}$ for $0 \leq x < 1$ . Its limit as $x ↑ 1$ is $1 / 4$ .

L1algebra

1.2

The inclusive partial sums satisfy $S_{2 m} = ι (m + 1)$ and $S_{2 m + 1} = - ι (m + 1)$ . Hence $σ_{2 m} = ι (m + 1) / ι (2 m + 1) \to 1 / 2$ , while $σ_{2 m + 1} = 0$ .

L2algebra

2.1

Thus the Cesaro means have two distinct subsequential limits and do not converge, whereas step 1.1 proves Abel summability to $1 / 4$ .

step 1.1step 1.2∎

Power Series and Real-Analytic Functions — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The geometric series represents 1/(1−x)1/(1−x) for ∣x∣<1∣x∣<1 and re-expands explicitly about every cc with ∣c∣<1∣c∣<1

Statement

Facts & Assumptions

Verification

The radius-one series with coefficients 1/(n+1)21/(n+1)2, 1/(n+1)1/(n+1) and 11 realise absolute, conditional and divergent endpoint behaviour

Statement

Facts & Assumptions

Verification

The series ∑n≥0ι(n!)xn∑n≥0​ι(n!)xn converges only at x=0x=0 and has radius zero

Statement

Facts & Assumptions

Verification

The geometric series converges pointwise but not uniformly on (−1,1)(−1,1)

Statement

Facts & Assumptions

Verification

A rational function with nonvanishing denominator is locally represented by geometric-series expansions

Statement

Facts & Assumptions

Verification

The alternating harmonic series illustrates Abel's boundary-limit theorem without evaluating its sum

Statement

Facts & Assumptions

Verification

Grandi's series is Abel summable to 1/21/2 but its partial sums do not converge

Statement

Facts & Assumptions

Verification

1−2+3−4+⋯1−2+3−4+⋯ is Abel summable to 1/41/4 but is not Cesaro summable

Statement

Facts & Assumptions

Verification

Sources

The geometric series represents $1 / (1 - x)$ for $∣ x ∣ < 1$ and re-expands explicitly about every $c$ with $∣ c ∣ < 1$

The radius-one series with coefficients $1 / (n + 1)^{2}$ , $1 / (n + 1)$ and $1$ realise absolute, conditional and divergent endpoint behaviour

The series $\sum_{n \geq 0} ι (n!) x^{n}$ converges only at $x = 0$ and has radius zero

The geometric series converges pointwise but not uniformly on $(- 1, 1)$

Grandi's series is Abel summable to $1 / 2$ but its partial sums do not converge

$1 - 2 + 3 - 4 + \dots$ is Abel summable to $1 / 4$ but is not Cesaro summable