LemmaStatement: AI-adaptedProof: AI-generatedverified 2026-09-26 (gpt-6-sol)

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The sine and cosine power series converge absolutely for every real argument

Statement

For every real $x$ , both series $\sum_{n = 0}^{\infty} \frac{(- 1)^{n} x^{2 n + 1}}{(2 n + 1)!} and \sum_{n = 0}^{\infty} \frac{(- 1)^{n} x^{2 n}}{(2 n)!}$ converge absolutely. Equivalently, both power series have infinite radius of convergence.

Facts & Assumptions

Given: A real $x$ .

[L1]

The two displayed series are real power series with the absolute terms $∣ x ∣^{2 n + 1} / (2 n + 1)!$ and $∣ x ∣^{2 n} / (2 n)!$ ; factorial denominators are positive (A real power series about a centre, its interval of convergence, and its radius in $[0, + \infty]$ , Integer powers $a^{m}$ , The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ , Canonical naturals are positive and strictly increasing).

[L2]

The ratio test proves absolute convergence when the ratio of successive absolute terms tends to a limit less than one (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence).

Proof

technique · direct

1.1L1algebra

If $x = 0$ , the sine series is identically zero and the cosine series has only its initial term $1$ ; both converge absolutely.

1.2L1algebra

If $x \neq 0$ , every absolute term is nonzero. The successive sine-term ratio is $∣ x ∣^{2} / ((2 n + 2) (2 n + 3))$ , and the successive cosine-term ratio is $∣ x ∣^{2} / ((2 n + 1) (2 n + 2))$ ; both tend to $0$ .

2.1step 1.1step 1.2L2∎

For $x \neq 0$ , the ratio test proves absolute convergence of both series. Together with step 1.1 this covers every real $x$ .

The sine and cosine power series converge absolutely for every real argument

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources