CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)

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The double sequence $(m + 1) / (m + n + 2)$ has unequal iterated limits

Statement refuted

Refuted claim: whenever both iterated limits of a double real sequence exist, they are equal.

For $m, n \in N$ define

$s_{m, n} : = \frac{ι (m + 1)}{ι (m + n + 2)} = \frac{ι (m + 1)}{ι (m + 1) + ι (n + 1)} .$

Then

$\lim_{m} \lim_{n} s_{m, n} = 0, \lim_{n} \lim_{m} s_{m, n} = 1.$

The shifts make the expression defined at the first index $(m, n) = (0, 0)$ .

Facts & Assumptions

Given: The double sequence $s_{m, n}$ in the Statement.

[L1]

The canonical-natural map satisfies $ι (0) = 0$ and $ι (r + 1) = ι (r) + 1$ ; positive canonical naturals increase, and their reciprocals decrease (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

For every real $ε > 0$ there is $N \geq 1$ with $1 / ι (N) < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L3]

A real sequence converges when its terms are eventually within every positive error of the proposed limit (Limits and Cauchy sequences of reals).

Counterexample

technique · direct

1.1

The denominator $ι (m + 1) + ι (n + 1)$ is positive for all $m, n$ , so every $s_{m, n}$ is defined and lies between $0$ and $1$ .

L1algebra

1.2

Fix $m$ and put $a : = ι (m + 1) > 0$ . Then $0 \leq s_{m, n} \leq a / ι (n + 1)$ ; given $ε > 0$ , [L2] makes the latter smaller than $ε$ for all sufficiently large $n$ . Hence $\lim_{n} s_{m, n} = 0$ .

L1L2L3choosealgebra

1.3

Fix $n$ and put $b : = ι (n + 1) > 0$ . Since $∣ 1 - s_{m, n} ∣ = b / (ι (m + 1) + b) \leq b / ι (m + 1)$ , [L2] makes this smaller than any prescribed $ε > 0$ for all sufficiently large $m$ . Hence $\lim_{m} s_{m, n} = 1$ .

L1L2L3choosealgebra

2.1

By step 1.2 the first inner-limit sequence is constantly $0$ , so its limit in $m$ is $0$ ; by step 1.3 the other inner-limit sequence is constantly $1$ , so its limit in $n$ is $1$ .

step 1.2step 1.3L3

3.1

Thus both iterated limits exist and are unequal, refuting the claim.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Stephen Abbott, Understanding Analysis, 2nd ed., Exercise 2.3.13 (standard reference, not scraped)

The double sequence (m+1)/(m+n+2)(m+1)/(m+n+2) has unequal iterated limits