Main derivation
S01 · definition
Before
u(n)\to 2
After
\forall\varepsilon>0\ \exists N\in\mathbb{N}_0\ \forall n\ge N:\ |u(n)-2|<\varepsilonConditions: none
Rule: mko_seq_limit · Uses: seq_limit_def, mko_seq_limit
Leaf basis: seq_limit_def + def_unfold
Unfold u(n)→2 by the MKO's definition. The quantifier order is 'for every ε there is N such that for all n≥N': N is chosen after ε and may depend on it; once chosen it must control the whole tail, and the error condition is the strict '<'.
S02 · definition
Before
u(n)=L+\frac{c}{n+d}After
L=2,\ c=-3,\ d=1,\ n+1\ge 1>0\ (n\in\mathbb{N}_0)Conditions: d\ge 1
Rule: mko_seq_limit · Uses: G1, G2
Leaf basis: def_intro + order_basics
Place the example in the MKO model: L=2, c=−3, d=1. Since n≥0, the denominator n+1≥1>0, so every term is defined and the denominator is positive (needed later when inequalities are inverted).
S03 · equivalent
Before
e(n)=|u(n)-2|
After
\left|\frac{-3}{n+1}\right|=\frac{3}{n+1}Conditions: n+1>0
Rule: abs_rules · Uses: S02, abs_rules
Leaf basis: abs_rules + order_basics + frac_rules
u(n)−2=−3/(n+1). Since n+1>0, 3/(n+1)>0, so −3/(n+1)<0 and by the definition of absolute value |−3/(n+1)|=3/(n+1). The error depends only on |c|=3 and n, not on L=2 (N5 spot-checks n=0 to 60).
S04 · implies
Before
n\ge N\ (n,N\in\mathbb{N}_0)After
\frac{3}{n+1}\le\frac{3}{N+1}Conditions: 0<N+1\le n+1
Rule: L_recip_mono · Uses: S02, S03
Leaf basis: L_recip_mono + order_basics
n≥N gives 0<N+1≤n+1. By Lemma L_recip_mono (a=N+1, b=n+1), 1/(n+1)≤1/(N+1); multiplying by the positive 3 keeps the inequality. So in this model the error does not increase with n: the largest error in the tail is at n=N.
S05 · equivalent
Before
\forall n\ge N:\ \frac{3}{n+1}<\varepsilonAfter
\frac{3}{N+1}<\varepsilonConditions: \varepsilon>0
Rule: L_cutoff · Uses: S03, S04
Leaf basis: forall_elim + forall_intro + order_basics + iff_intro
By Lemma L_cutoff, a candidate N controls the whole tail exactly when the N-th error itself is below ε. This turns MKO assumption a2, 'a uniform cutoff reduces to one exact inequality', into a proof. The lemma is proved in S05.1 and S05.2.
S06 · equivalent
Before
\frac{3}{N+1}<\frac{p}{q}After
3q<p(N+1)
Conditions: q>0; N+1>0
Rule: order_basics · Uses: S05, G3, frac_rules
Leaf basis: order_basics + frac_rules + mul_inverse_nonzero
Multiply both sides by the positive q(N+1); the direction is unchanged: 3q<p(N+1). This is the MKO's exact condition |c|q<p(N+d), involving integers only. The strict '<' is kept: equality does not qualify.
S07 · equivalent
Before
3\cdot 2<1\cdot(N+1)
After
N\ge 6\ (N\in\mathbb{N}_0)Conditions: N\ \text{integer}
Rule: order_basics · Uses: S06, G3
Leaf basis: order_basics + add_inverse + numeric_evaluation
Substitute p=1, q=2: 6<N+1, that is N>5. N is an integer and there is no integer strictly between 5 and 6, so N>5 ⇔ N≥6.
S08 · implies
Before
N=5
After
e(5)=\frac{3}{6}=\frac12\not<\frac12:\ N=5\ \text{invalid}Conditions: none
Rule: L_cutoff · Uses: S03, S05, G3
Leaf basis: L_cutoff + contrapositive + numeric_evaluation
For N=5 the error of term 5 is 3/6=1/2, exactly ε, not below ε; by the contrapositive of the ⇒ direction of S05, N=5 is not a valid cutoff. This is the MKO's warning that an error equal to ε is not less than ε.
S09 · implies
Before
N=6:\ \frac{3}{7}<\frac12After
\forall n\ge 6:\ e(n)<\frac12;\ \text{least valid}\ N=6Conditions: none
Rule: L_cutoff · Uses: S05, S07, S08
Leaf basis: L_cutoff + order_basics + numeric_evaluation
3·2=6<7=1·(6+1), so 3/7<1/2; by S05 (⇐), every n≥6 has error below 1/2. By S07 the valid cutoffs are exactly N≥6, so the least valid N is 6, agreeing with the MKO formula max(0, floor(3·2/1)−1+1)=6 (N4). N6 is a finite spot check for n=6 to 200, not a proof; the proof is S05.
S10 · implies
Before
\varepsilon>0\ \text{arbitrary}After
N\in\mathbb{N}_0,\ N>\frac{3}{\varepsilon},\ \text{hence}\ \frac{3}{N+1}<\varepsilonConditions: \varepsilon>0
Rule: archimedean · Uses: archimedean, S01
Leaf basis: archimedean + order_basics + frac_rules
Take an arbitrary ε>0 with no other condition. 3/ε is a real number, so by the Archimedean property there is a natural number N>3/ε, and then N+1>3/ε>0. Multiplying both sides by the positive ε/(N+1) gives ε>3/(N+1). For example, for ε=1/10 one may take N=31 (N12); N need not be the least one.
S11 · implies
Before
\frac{3}{N+1}<\varepsilonAfter
\forall n\ge N:\ |u(n)-2|=\frac{3}{n+1}<\varepsilonConditions: none
Rule: L_cutoff · Uses: S10, S05, S03
Leaf basis: L_cutoff + eq_replace
By the ⇐ direction of S05, the N chosen in S10 controls the whole tail; by S03, 3/(n+1) is exactly the error |u(n)−2|.
S12 · implies
Before
\varepsilon\ \text{arbitrary},\ \exists N\ \forall n\ge N:\ |u(n)-2|<\varepsilonAfter
u(n)\to 2
Conditions: none
Rule: forall_intro · Uses: S10, S11, S01
Leaf basis: forall_intro + exists_intro + seq_limit_def
S10 and S11 construct a valid N for an arbitrary ε>0, with no extra condition on ε; by generalization, every ε>0 has a valid N. By the definition in S01, u(n)→2. The proof covers every ε; it does not merely test a few values of ε or a few terms.
S13 · implies
Before
\exists N\ \forall\varepsilon>0\ \forall n\ge N:\ e(n)<\varepsilon\ ?
After
\text{false: for each}\ N,\ \varepsilon=\tfrac{3}{N+1}\ \text{gives}\ e(N)=\varepsilon\not<\varepsilonConditions: none
Rule: quantifier_negation · Uses: S03, S05, S06, G3
Leaf basis: quantifier_negation + L_cutoff + numeric_evaluation
Fixing N before ε turns the statement into 'one N is valid for every ε'. That is false: for any N take ε=3/(N+1)>0; the N-th error equals ε exactly and is not below it. Concretely, the least N for ε=1/2 is 6, but for ε=1/10, 3/7>1/10, so N=6 fails; by S06, ε=1/10 needs 3·10<N+1, that is N≥30 (at N=29 the error is exactly 1/10). N must change with ε.
Error controls and negative controls
An error equal to ε is not below ε
One misconception the MKO lists is treating an error equal to ε as below ε. For ε=1/2 the error of term 5 is 3/6=1/2 exactly, so N=5 is invalid (S08), and the least N is 6, not 5. The MKO's own example c=1, d=1, ε=1/2 is the same: at N=1 the error 1/2 equals the tolerance and the least N is 2 (N13, N14).
N is chosen after ε
Freezing the N that comes after ∀ε is another misconception. S13 proves that no single N works for every ε: N=6 suffices for ε=1/2, but ε=1/10 needs N≥30. The correct quantifier order is ∀ε∃N (S12), not ∃N∀ε.
Finitely many terms do not determine a limit
These finitely many terms do not determine convergence of the whole sequence to the specified value 2: the extension v below converges to 5 but not to 2. Let v(n)=u(n) for n≤100 and v(n)=5 for n>100: its first 101 terms equal those of u, but with ε=1 and any N, taking n=max(N,101) gives |v(n)−2|=3≥1, so v does not tend to 2. Here checking the N-th term suffices only because S04 proved this model's errors do not increase; a general sequence has no such guarantee.
- For ε=1/2, N=5 is valid because the error of term 5 equals 1/2 →
reject — The condition is the strict |u(n)−2|<ε; 3/6=1/2 is not below 1/2. - N=6 is valid for every ε>0 →
reject — For ε=1/10 the error of term 6 is 3/7>1/10. - The first 101 terms of v(n) equal those of u(n), so v(n)→2 →
reject — v(n)=5 for n>100; with ε=1 every N fails at n=max(N,101). - Changing L=2 to L=100 changes the least cutoff for ε=1/2 →
reject — The error |c|/(n+d) does not involve L (S03); the least N is still 6.