TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-07-31

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Abel's limit theorem: if a real series converges to $s$ , then its power series tends to $s$ as $x ↑ 1$

Statement

If the real series $\sum_{n \geq 0} a_{n}$ converges ordinarily to $s$ , then it is Abel summable to $s$ :

$\lim_{x ↑ 1} \sum_{n = 0}^{\infty} a_{n} x^{n} = s .$

Facts & Assumptions

Given: Inclusive partial sums $S_{n} : = \sum_{k = 0}^{n} a_{k}$ with $S_{n} \to s$ .

[L1]

Finite Abel summation gives $\sum_{k = 0}^{N} a_{k} x^{k} = S_{N} x^{N} + (1 - x) \sum_{k = 0}^{N - 1} S_{k} x^{k}$ (Abel summation by parts: with $A_{n} = \sum_{k < n} a_{k}$ one has $\sum_{k < n} a_{k} b_{k} = A_{n} b_{n - 1} - \sum_{k < n - 1} A_{k + 1} (b_{k + 1} - b_{k})$ for every $n \geq 1$ ).

[L2]

A convergent sequence is bounded (Every convergent sequence is bounded).

[L3]

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

[L4]

A nonnegative series dominated termwise by a convergent nonnegative series converges (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ ).

Proof

technique · direct

1.1L2L3L4choose

By [L2], choose $M$ with $∣ S_{n} ∣ \leq M$ for every $n$ . For fixed $0 \leq x < 1$ , one has $∣ S_{n} x^{n} ∣ \leq M x^{n}$ , so [L3] and [L4] give absolute convergence of $\sum_{n} S_{n} x^{n}$ ; the same bound gives $S_{N} x^{N} \to 0$ .

2.1step 1.1L1L3

Apply [L1] and let $N \to \infty$ . Step 1.1 gives convergence of the Abel series and $A (x) = (1 - x) \sum_{n \geq 0} S_{n} x^{n}$ . Subtracting $s = (1 - x) \sum_{n \geq 0} s x^{n}$ gives $A (x) - s = (1 - x) \sum_{n \geq 0} (S_{n} - s) x^{n}$ .

3.1givenstep 2.1L3choose

Given $ε > 0$ , choose $N$ so that $∣ S_{n} - s ∣ < ε$ for $n \geq N$ . The tail of step 2.1 has absolute value at most $ε (1 - x) \sum_{n \geq N} x^{n} \leq ε$ .

4.1step 2.1step 3.1∎

The finite head $(1 - x) \sum_{n < N} (S_{n} - s) x^{n}$ tends to $0$ as $x ↑ 1$ . Thus $∣ A (x) - s ∣ < 2 ε$ for all sufficiently large $x < 1$ , proving the asserted one-sided limit and Abel summability.

Abel's limit theorem: if a real series converges to $s$ , then its power series tends to $s$ as $x ↑ 1$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Abel's limit theorem: if a real series converges to ss, then its power series tends to ss as x↑1x↑1

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Abel's limit theorem: if a real series converges to $s$ , then its power series tends to $s$ as $x ↑ 1$