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CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31
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The geometric series converges pointwise but not uniformly on (−1,1)(-1,1)

Statement

The partial sums of ∑n≥0xn\sum_{n\ge0}x^n converge pointwise to 1/(1−x)1/(1-x) on (−1,1)(-1,1) but do not converge uniformly there.

Facts & Assumptions

Given: The geometric-series partial sums on (−1,1)(-1,1).

[L1]

Pointwise convergence follows from the geometric-series theorem (For ∣r∣<1|r| < 1, ∑k≥0rk=1/(1−r)\sum_{k \ge 0} r^k = 1/(1-r), and for ∣r∣≥1|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 xNx^N, whose supremum over (−1,1)(-1,1) is 11 for every NN.

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∎

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