derivation-mko-monotone-sequence-convergence-theorem-01 · ocme-new142-reader-adoption-20261021
For a_n=3-2/(n+1), prove monotonicity, boundedness, supremum 3 and convergence.
A full n≥0 law with a strict epsilon certificate; explicitly include the general theorem's supremum step.
algebraepsilonsupremumdefinition_hereG1 — a_n=3-2/(n+1) for every n≥0; use the least-upper-bound property of R.a_n — concrete whole sequences — general theorem supremumSome rule statements have reader clarifications in this revision: they are paraphrases, not verbatim quotations. Referenced original source bytes and hashes are unchanged.
mko_definition · cited — Keep whole-sequence monotonicity and the corresponding bound. If s is the supremum of the terms, s−epsilon is not an upper bound, supplying an index N; monotonicity then controls every later term. A supremum or infimum need not be a term. A finite prefix does not supply a universal certificate. A monotone sequence without the required bound, or a bounded sequence lacking monotonicity, does not meet this theorem's hypotheses. · mko-monotone-sequence-convergence-theoremalgebra · prerequisite — Use real-field identities; division requires a nonzero denominator and multiplication by a positive number preserves inequalities.epsilon · prerequisite — Convergence requires for every positive epsilon an integer N controlling every n≥N by a strict error bound; use the Archimedean property of the reals.supremum · prerequisite — A nonempty bounded-above real set has supremum s; for epsilon>0, s-epsilon is not an upper bound.definition_here · prerequisite — The epsilon definition of real-sequence limit L requires, for each positive epsilon, an N with |u_n-L|<epsilon for every n≥N.Before
a_{n+1}-a_n=2/(n+1)-2/(n+2)After
a_{n+1}-a_n=2/((n+1)(n+2))>0Conditions: n\ge0
Rule: algebra · Uses: G1
Leaf basis: algebra
Both denominators are positive, proving every adjacent difference positive and hence whole-sequence strict increase.
Before
0<2/(n+1)\le2
After
1\le a_n<3
Conditions: none
Rule: algebra · Uses: G1
Leaf basis: algebra
This gives lower bound 1 and the upper bound 3 needed for the increasing-sequence theorem.
Before
\varepsilon>0,\quad N=\lfloor2/\varepsilon\rfloor+1
After
0<3-a_n=2/(n+1)<\varepsilon\quad(n\ge N)
Conditions: none
Rule: epsilon · Uses: G1, S02
Leaf basis: epsilon
Every later denominator is larger, directly proving a_n tends to 3.
Before
u<3,\quad\varepsilon=3-u>0
After
a_N>3-\varepsilon=u;\quad\sup\{a_n:n\ge0\}=3Conditions: none
Rule: supremum · Uses: S02, S03
Leaf basis: supremum
Every number below 3 fails as an upper bound; combined with the upper-bound check this gives leastness.
Before
(u_n)\text{ nondecreasing and bounded above},\quad s=\sup\{u_n\}After
\forall\varepsilon>0\ \exists N:\ s-\varepsilon<u_N\le s
Conditions: (u_n)\text{ nondecreasing and bounded above}
Rule: supremum · Uses: G1
Leaf basis: supremum
For the general theorem's core, s-epsilon is not an upper bound, so some term exceeds it.
Before
n\ge N
After
s-\varepsilon<u_N\le u_n\le s
Conditions: none
Rule: algebra · Uses: S05
Leaf basis: algebra
Whole-sequence monotonicity transfers one close term to the entire tail.
Before
0\le s-u_n<\varepsilon\quad(n\ge N)
After
u_n\to s
Conditions: none
Rule: definition_here · Uses: S06
Leaf basis: definition_here
The absolute error equals s-u_n, meeting the limit definition and completing the supremum proof.
Before
a_n\text{ nondecreasing},\quad\sup\{a_n\}=3After
\lim a_n=3\text{, not attained by any }a_nConditions: none
Rule: epsilon · Uses: S01, S02, S04, S07
Leaf basis: epsilon
The general proof and direct tail agree on the limit, retaining the nonattainment boundary.
No substeps in this guide.
S08 · premise_audit · S01, S02, S04, S07 — No converse claiming all convergent sequences monotone is inferred.S07 · premise_audit · S06 — Only the stated monotonicity and bound premises are used, not inferred from samples.S06 · premise_audit · S05 — Without monotonicity, u_N cannot control all later terms.S05 · premise_audit · G1 — This separately names a general u_n and both premises, avoiding symbol collision with a_n.S04 · premise_audit · S02, S03 — A supremum need not be a term; no term equals 3 here.S03 · premise_audit · G1, S02 — N is fixed once for the entire later tail.S02 · premise_audit · G1 — The upper bound is not a maximum because every term is strictly below 3.S01 · premise_audit · G1 — The formula holds for every n, not sampled neighbors.Pair nondecreasing with an upper bound and nonincreasing with a lower bound.
Failure of s-epsilon as an upper bound supplies a strictly close term.
reject — Every term is strictly below 3.reject — n increases without an upper bound and has no finite limit.reject — (-1)^n is bounded with different parity limits.a_n is strictly increasing and bounded, tends to 3, and has unattained supremum 3.
1 declared checks for this guide; only the listed expressions are covered.
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