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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 $|x|<1$ ,

$\frac1{1-x}=\sum_{n=0}^{\infty}x^n.$

More generally, if $|c|<1$ , then

$\frac1{1-x}=\sum_{n=0}^{\infty}\frac{(x-c)^n}{(1-c)^{n+1}}\qquad(|x-c|<1-c).$

Facts & Assumptions

Given: A real $c$ with $|c|<1$ .

[L1]

The geometric series sums to $1/(1-t)$ for $|t|<1$ (For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ 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

$\sum_{n\ge0}\frac{x^n}{\iota(n+1)^2},\qquad \sum_{n\ge0}\frac{x^n}{\iota(n+1)},\qquad \sum_{n\ge0}x^n$

has radius $1$ . At $x=\pm1$ 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 $\iota(n)^{1/n}\to1$ and limits respect products and reciprocals ( $n^{1/n} \to 1$ , The canonical natural $\iota(n) = n \cdot 1_F$ 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 $\limsup_{k\to\infty}|a_{k+1}|^{1/(k+1)}$ , with the zero and infinite cases included).

[L3]

The $p$ -series converges for rational $p>1$ and diverges for $p\le1$ , 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 $\sum_{n\ge0}\iota(n!)x^n$ converges only at $x=0$ and has radius zero

Statement

The real power series $\sum_{n\ge0}\iota(n!)x^n$ has radius $0$ : it converges at its centre $x=0$ and diverges at every $x\ne0$ .

Facts & Assumptions

Given: Coefficients $a_n:=\iota(n!)$ .

[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 $\mathbb{N}$ , The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

[L2]

Cauchy–Hadamard assigns radius $0$ when the coefficient-root limit superior is $+\infty$ (Cauchy–Hadamard: the reciprocal radius is $\limsup_{k\to\infty}|a_{k+1}|^{1/(k+1)}$ , 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 $\iota((k+1)!)^{1/(k+1)}>M$ 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 $x\ne0$ 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 $\sum_{n\ge0}x^n$ 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 $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ 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 $\mathbb R\setminus\{2\}$ . At every $c\ne2$ it has the local expansion

$\frac1{2-x}=\sum_{n=0}^{\infty}\frac{(x-c)^n}{(2-c)^{n+1}}\qquad(|x-c|<|2-c|).$

Facts & Assumptions

Given: A centre $c\ne2$ .

[L1]

The geometric series converges for arguments of absolute value below $1$ (For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ 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 $c\ne2$ 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

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

Then, without evaluating $s$ ,

$\lim_{x\uparrow1}\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{\iota(n+1)}=s.$

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 $\iota(n) = n \cdot 1_F$ 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 $x\uparrow1$ ).

Verification

technique · direct

1.1

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

2.1

Apply [L2] to its coefficients $a_n=(-1)^n/\iota(n+1)$ 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 $1-1+1-1+\cdots$ is Abel summable to $1/2$ but diverges ordinarily.

Facts & Assumptions

Given: Coefficients $a_n=(-1)^n$ .

[L1]

For $0\le x<1$ , $\sum_{n\ge0}(-1)^nx^n=1/(1+x)$ (For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ the series diverges).

[L2]

Abel summability and ordinary convergence are defined through the boundary limit and partial sums, respectively (Abel summability by $\lim_{x\uparrow1}\sum a_nx^n$ 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 $x\uparrow1$ .

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 →

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

Statement

The series $\sum_{n\ge0}(-1)^n\iota(n+1)$ is Abel summable to $1/4$ , but its Cesaro means do not converge.

Facts & Assumptions

Given: Coefficients $a_n=(-1)^n\iota(n+1)$ .

[L1]

The geometric series for $1/(1+x)$ may be differentiated term by term for $|x|<1$ (For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ 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 $\sigma_n = (x_0 + \dots + x_n)/(n+1)$ and $(C,1)$ -summability, Abel summability by $\lim_{x\uparrow1}\sum a_nx^n$ and Cesaro summability by the Cesaro means of the partial sums).

Verification

technique · direct

1.1

Differentiating $\sum_{n\ge0}(-1)^nx^n=1/(1+x)$ and combining with the original series gives $\sum_{n\ge0}(-1)^n\iota(n+1)x^n=1/(1+x)^2$ for $0\le x<1$ . Its limit as $x\uparrow1$ is $1/4$ .

L1algebra

1.2

The inclusive partial sums satisfy $S_{2m}=\iota(m+1)$ and $S_{2m+1}=-\iota(m+1)$ . Hence $\sigma_{2m}=\iota(m+1)/\iota(2m+1)\to1/2$ , while $\sigma_{2m+1}=0$ .

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∎

Sources

Standard references

Recommended treatments; not extraction sources.

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)1/(1−x) for ∣x∣<1|x|<1∣x∣<1 and re-expands explicitly about every ccc with ∣c∣<1|c|<1∣c∣<1

Statement

Facts & Assumptions

Verification

The radius-one series with coefficients 1/(n+1)21/(n+1)^21/(n+1)2, 1/(n+1)1/(n+1)1/(n+1) and 111 realise absolute, conditional and divergent endpoint behaviour

Statement

Facts & Assumptions

Verification

The series ∑n≥0ι(n!)xn\sum_{n\ge0}\iota(n!)x^n∑n≥0​ι(n!)xn converges only at x=0x=0x=0 and has radius zero

Statement

Facts & Assumptions

Verification

The geometric series converges pointwise but not uniformly on (−1,1)(-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 1/21/21/2 but its partial sums do not converge

Statement

Facts & Assumptions

Verification

1−2+3−4+⋯1-2+3-4+\cdots1−2+3−4+⋯ is Abel summable to 1/41/41/4 but is not Cesaro summable

Statement

Facts & Assumptions

Verification

Sources

Standard references

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 $\sum_{n\ge0}\iota(n!)x^n$ 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

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