CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-07-31

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A real power series converges absolutely inside its radius and diverges outside it, while either behaviour may occur at an endpoint

Statement

Let $\sum a_{n} (x - c)^{n}$ have radius $R$ . It converges absolutely at every $x$ with $∣ x - c ∣ < R$ and diverges at every $x$ with $∣ x - c ∣ > R$ . When $0 < R < + \infty$ , no common conclusion holds at either endpoint $c \pm R$ : power series of radius $R$ can converge there, even absolutely, or diverge there.

Facts & Assumptions

Given: A real power series $\sum a_{n} (x - c)^{n}$ with radius $R$ (A real power series about a centre, its interval of convergence, and its radius in $[0, + \infty]$ ).

[L1]

Cauchy-Hadamard identifies $R$ from the limit superior of the coefficient roots and the root test gives absolute convergence below the reciprocal threshold and divergence above it (Cauchy–Hadamard: the reciprocal radius is ${lim sup}_{k \to \infty} ∣ a_{k + 1} ∣^{1 / (k + 1)}$ , with the zero and infinite cases included).

[L2]

At root-test boundary value $1$ , the coefficient families $1 / (n + 1)$ and $1 / (n + 1)^{2}$ both have root limit superior $1$ , while the first series diverges and the second converges; changing the coefficient signs does not change their absolute values (Root test: $lim sup ∣ a_{k} ∣^{1 / k} < 1$ gives absolute convergence and hence convergence, $> 1$ gives divergence, and $= 1$ decides nothing, claim 3).

Proof

technique · direct

1.1L1

The assertions for $∣ x - c ∣ < R$ and $∣ x - c ∣ > R$ are exactly the two strict alternatives supplied by [L1], including the cases $R = 0$ and $R = + \infty$ .

1.2L2

For endpoint behaviour at radius $1$ , the series with coefficients $1 / (n + 1)^{2}$ converges absolutely at both $x = 1$ and $x = - 1$ . The series with coefficients $1 / (n + 1)$ diverges at $x = 1$ , while the series with coefficients $(- 1)^{n} / (n + 1)$ diverges at $x = - 1$ . All three have radius $1$ by [L2].

2.1step 1.2algebra∎

Replacing $x$ by $(x - c) / R$ and multiplying coefficients by the corresponding powers of $R^{- 1}$ transports the two radius-one examples to any finite $R > 0$ and centre $c$ . Thus either behaviour may occur at an endpoint, while no assertion has been made when the endpoints are not real.

A real power series converges absolutely inside its radius and diverges outside it, while either behaviour may occur at an endpoint

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources