Questions: Uniform Convergence of Power Series

Question 1

A power series has radius of convergence R = 3. On which domain is term-by-term integration rigorously justified by the uniform convergence theorem?

• A. The open interval (−3, 3), since the series converges at every point there
• B. The closed interval [−3, 3], which is the full interval of convergence
• C. Any closed interval [−r, r] with r < 3, where uniform convergence is guaranteed by the Weierstrass M-test
• D. Only at the center x = 0, where all terms are trivially bounded

Answer: (C)

Uniform convergence (and the resulting validity of term-by-term operations) holds on any closed interval strictly inside the radius of convergence. On a closed [−r, r] with r < R, each term |aₙ(x−c)ⁿ| ≤ |aₙ|rⁿ = Mₙ, and ΣMₙ converges. On the full open interval (option A), the bound breaks down near the boundary. On [−R, R] (option B), convergence at the endpoints may be conditional, and uniform convergence can fail there.

Question 2

In applying the Weierstrass M-test to show ∑aₙxⁿ converges uniformly on [−r, r] (with r < R), what is the bounding sequence Mₙ, and why must r be strictly less than R?

• A. Mₙ = |aₙ| — the coefficients alone bound the terms
• B. Mₙ = |aₙ|Rⁿ — using the full radius ensures the tightest possible bound
• C. Mₙ = |aₙ|rⁿ — this dominates all terms on [−r, r] and ΣMₙ converges because r < R
• D. Mₙ = n|aₙ|rⁿ⁻¹ — the derivative series provides the correct bound

Answer: (C)

For any x in [−r, r], |aₙxⁿ| ≤ |aₙ|rⁿ = Mₙ. The M-test requires ΣMₙ < ∞. Since r < R, the series ∑|aₙ|rⁿ converges (this is what having a radius of convergence means). If we tried r = R (option B), ∑|aₙ|Rⁿ might diverge, and the M-test would fail. Using just |aₙ| (option A) ignores the x-dependence entirely and doesn't give a valid bound for all n.

Question 3

A power series that converges pointwise at most point of (−R, R) necessarily converges uniformly on that entire open interval.

• T. True
• F. False

Answer: False

Pointwise convergence on an open interval does not imply uniform convergence on that same interval. The theorem guarantees uniform convergence only on closed sub-intervals [−r, r] with r strictly less than R. Near the endpoints, the partial sums may converge more and more slowly, preventing uniformity on the full open interval. This is precisely why the Weierstrass M-test is applied to a compact sub-interval: the bound Mₙ = |aₙ|rⁿ is only summable because r < R.

Question 4

Term-by-term differentiation of a power series ∑aₙ(x − c)ⁿ produces a new power series with the same radius of convergence as the original.

• T. True
• F. False

Answer: True

The differentiated series is ∑naₙ(x − c)ⁿ⁻¹. The radius of convergence is determined by lim sup |aₙ|^(1/n) via the root test, and the factor of n does not affect this limit since n^(1/n) → 1. So the radius of convergence is unchanged. Similarly for integration. This means a power series can be differentiated and integrated infinitely many times within its disk of convergence — making analytic functions the most well-behaved class in analysis.

Question 5

Why must you restrict to a closed sub-interval strictly inside the interval of convergence to establish uniform convergence via the Weierstrass M-test? What goes wrong at the boundary of the interval of convergence?

Model answer: The M-test requires finding Mₙ such that |aₙ(x−c)ⁿ| ≤ Mₙ for all x in the interval and ΣMₙ < ∞. On [−r, r] with r < R, the bound Mₙ = |aₙ|rⁿ works because ∑|aₙ|rⁿ converges (by definition of R). At the boundary x = R, the bound would need Mₙ = |aₙ|Rⁿ, but ∑|aₙ|Rⁿ may diverge — the series might only converge conditionally there, not absolutely. Without an absolutely convergent dominating series, the M-test cannot be applied.

The boundary behavior of a power series is delicate: it can converge absolutely, converge conditionally, or diverge at x = ±R, depending on the specific series. Uniform convergence on an open interval would require controlling the supremum over all points, including those approaching R — and near R the partial sums may converge arbitrarily slowly. The compactness of [−r, r] with r < R is what lets the geometric bound close.