Questions: Weierstrass M-Test

Question 1

The series ∑_{n=1}^∞ sin(nx)/n² is being analyzed for uniform convergence on all of ℝ. Which of the following correctly applies the Weierstrass M-test?

• A. Since |sin(nx)/n²| ≤ 1/n² for all x ∈ ℝ and ∑1/n² = π²/6 converges, the M-test gives uniform convergence on ℝ
• B. Since sin(nx)/n² → 0 pointwise for each x, uniform convergence follows automatically
• C. The series converges uniformly because the partial sums are uniformly bounded
• D. The M-test cannot be applied because sin(nx) oscillates and has no fixed bound

Answer: (A)

The M-test requires finding constants Mₙ ≥ |fₙ(x)| for all x simultaneously. Here Mₙ = 1/n² works because |sin(nx)| ≤ 1 for all x, so |sin(nx)/n²| ≤ 1/n² regardless of x. Since ∑1/n² converges, the M-test applies. Option B is wrong because pointwise convergence (each x independently) does not imply uniform convergence — uniform convergence requires the same N to work for all x at once. Option D misunderstands the test: sin(nx) is bounded, and that bound is all we need.

Question 2

A student concludes: 'The series ∑fₙ(x) fails the Weierstrass M-test on [0,1] because no constants Mₙ with ∑Mₙ convergent can bound all |fₙ(x)|. Therefore the series does not converge uniformly on [0,1].' This reasoning is:

• A. Correct: the M-test is both necessary and sufficient for uniform convergence
• B. Incorrect: the M-test is sufficient but not necessary — failing it does not rule out uniform convergence by other methods
• C. Correct only when the functions fₙ are continuous on [0,1]
• D. Incorrect because the M-test always applies to series on closed bounded intervals

Answer: (B)

The Weierstrass M-test is a sufficient condition, not a necessary one. A series can converge uniformly on a set even when no valid sequence of dominating constants Mₙ exists with ∑Mₙ < ∞. The alternating series ∑(-1)ⁿ/n converges uniformly on certain sets despite failing the M-test. When the M-test fails, other tools — Dirichlet's test, Abel's test — may still establish uniform convergence. Failing the M-test only means this particular tool cannot help; it is not a proof of non-uniform convergence.

Question 3

The Weierstrass M-test establishes uniform convergence by finding x-independent constants Mₙ ≥ |fₙ(x)| for all x, and the critical feature is that these constants are independent of x — making the same bound hold everywhere simultaneously.

• T. True
• F. False

Answer: True

The x-independence of the constants Mₙ is exactly what converts the argument from pointwise to uniform. When ∑Mₙ < ∞, the tail ∑_{n>N} Mₙ can be made smaller than any ε by choosing N large enough — and because Mₙ ≥ |fₙ(x)| for *every* x, the same N makes the tail of the function series smaller than ε everywhere simultaneously. This is the definition of uniform convergence: a single N that works across all x, rather than an N that may depend on the specific x chosen.

Question 4

If a series of functions ∑fₙ(x) converges pointwise on a set S, the Weierstrass M-test can be applied to conclude it also converges uniformly.

• T. True
• F. False

Answer: False

Pointwise convergence and uniform convergence are genuinely different, and the M-test is not a bridge between them. Pointwise convergence means: for each fixed x, the partial sums eventually stay within ε of the limit — but the required N may depend on x. Uniform convergence requires a single N that works for all x. The M-test establishes uniform convergence directly (without going through pointwise convergence); it cannot be applied to series merely because they converge pointwise. The classic counterexample ∑_{n} fₙ(x) = x^n on [0,1] converges pointwise but not uniformly.

Question 5

Why is pointwise convergence insufficient for swapping limits with integration or differentiation, and how does the Weierstrass M-test restore these properties?

Model answer: Pointwise convergence allows the rate of convergence to vary arbitrarily across x — some points may require many more terms than others. When you integrate or differentiate, this x-dependent variation can accumulate or interact in ways that break the limit-integral interchange. Uniform convergence, by contrast, means the worst-case error over all x goes to zero, so the limit function is approached at a uniform rate. Under uniform convergence, ∫∑fₙ = ∑∫fₙ and (under additional conditions) d/dx ∑fₙ = ∑ d/dx fₙ. The M-test establishes uniform convergence by bounding each |fₙ(x)| by an x-independent Mₙ with ∑Mₙ < ∞, ensuring the tail of the series is uniformly small.

The practical payoff of the M-test is precisely these interchange theorems. Once you know ∑fₙ converges uniformly on S, you can integrate term by term, differentiate term by term (under mild additional conditions), and conclude the limit function is continuous if all fₙ are continuous. These conclusions fail for merely pointwise convergent series, where the limit function can be discontinuous even when all fₙ are continuous. The M-test is the most commonly invoked tool to access these powerful consequences.