CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $m,n\in\mathbb{N}$ define

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

Then

$\lim_m\lim_ns_{m,n}=0,\qquad \lim_n\lim_ms_{m,n}=1.$

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 $\iota(0)=0$ and $\iota(r+1)=\iota(r)+1$ ; positive canonical naturals increase, and their reciprocals decrease (The canonical natural $\iota(n) = n \cdot 1_F$ of a field, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

For every real $\varepsilon>0$ there is $N\ge1$ with $1/\iota(N)<\varepsilon$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ).

[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 $\iota(m+1)+\iota(n+1)$ 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 $a:=\iota(m+1)>0$ . Then $0\le s_{m,n}\le a/\iota(n+1)$ ; given $\varepsilon>0$ , [L2] makes the latter smaller than $\varepsilon$ for all sufficiently large $n$ . Hence $\lim_ns_{m,n}=0$ .

L1L2L3choosealgebra

1.3

Fix $n$ and put $b:=\iota(n+1)>0$ . Since $|1-s_{m,n}|=b/(\iota(m+1)+b)\le b/\iota(m+1)$ , [L2] makes this smaller than any prescribed $\varepsilon>0$ for all sufficiently large $m$ . Hence $\lim_ms_{m,n}=1$ .

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

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