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

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Inside the common radius the product of two power-series sums is represented by the Cauchy product of their coefficients

Statement

Suppose $f (x) = \sum_{n \geq 0} a_{n} (x - c)^{n}$ and $g (x) = \sum_{n \geq 0} b_{n} (x - c)^{n}$ have radii $R_{f}, R_{g}$ . For $∣ x - c ∣ < \min (R_{f}, R_{g})$ ,

$f (x) g (x) = \sum_{n = 0}^{\infty} (\sum_{k = 0}^{n} a_{k} b_{n - k}) (x - c)^{n},$

and the displayed product series converges absolutely.

Facts & Assumptions

Given: The two power series in the statement and a point in their common open radius.

[L1]

Both numerical series converge absolutely there (A real power series converges absolutely inside its radius and diverges outside it, while either behaviour may occur at an endpoint).

[L2]

The Cauchy product of two absolutely convergent series converges absolutely to the product of their sums (The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ , If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$ ).

Proof

technique · direct

1.1L1L2

Apply [L2] to the numerical series with terms $a_{k} (x - c)^{k}$ and $b_{j} (x - c)^{j}$ , whose absolute convergence is [L1].

2.1step 1.1algebra∎

Its $n$ th Cauchy-product term is $\sum_{k = 0}^{n} a_{k} b_{n - k} (x - c)^{n}$ , which gives the formula and absolute convergence.

Inside the common radius the product of two power-series sums is represented by the Cauchy product of their coefficients

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources