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Part BCSIR NET December 2024split-the-term-before-applying-the-m-test

Split the term before applying the m test

Consider the sequences (fn)n≥1(f_{n})_{n}\ge_{1} and (gn)n≥1(g_{n})_{n}\ge_{1} of functions defined on the interval [−1, 1] by fn(x)=(−1)n(x2+n)/n2f_{n}(x) = (-1)^{n}(x^{2} + n)/n^{2} and gn(x)=(−1)n(x2+n2)/n3g_{n}(x) = (-1)^{n}(x^{2} + n^{2})/n^{3}. Which of the following statements is true?

  1. A.∑(n≥1)fn\sum(n\ge1) f_{n} and ∑(n≥1)gn\sum(n\ge1) g_{n} are uniformly convergent on the interval [−1, 1].✓ Correct
  2. B.∑(n≥1)fn\sum(n\ge1) f_{n} is uniformly convergent on the interval [−1, 1], but ∑(n≥1)gn\sum(n\ge1) g_{n} is not.
  3. C.∑(n≥1)gn\sum(n\ge1) g_{n} is uniformly convergent on the interval [−1, 1], but ∑(n≥1)fn\sum(n\ge1) f_{n} is not.
  4. D.Neither ∑(n≥1)fn\sum(n\ge1) f_{n} nor ∑(n≥1)gn\sum(n\ge1) g_{n} is uniformly convergent on the interval [−1, 1].

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests pointwise vs uniform.

See pricing

50 are analysed free — try those first.

The trap it tests

Pointwise vs uniform

The mode of convergence — or of continuity — was the wrong one.

Drill statements like this

Related counterexample: fₙ → f uniformly ⇒ fₙ′ → f′

More on this topic

The chapter behind this: Uniform convergence — what it buys and how to test it — free to read

From Sequences and Series of Functions › Pointwise vs uniform convergence, M-test, Dini

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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