Definition D1
lim aₙ = L means the following. For every ε > 0 there exists a natural number N such that |aₙ − L| < ε for every n with n ≥ N.
Proof.
(1)ε is given first, and N is chosen afterward.convention
(2)This definition speaks only of the terms after N. It says nothing about where the terms with n < N are.convention
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Definition D2
The sequence (aₙ) is bounded means that there is a real number M with |aₙ| ≤ M for every natural number n.
Proof.
M is a single number that does not depend on n. Having, for each n, some number larger than |aₙ| is not boundedness.convention
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Theorem T1
If lim aₙ = L, then (aₙ) is bounded.
Proof.
(1)Let ε = 1. Then 0 < 1.A5
(2)For this ε there is a natural number N such that |aₙ − L| < 1 whenever n ≥ N.assumption, (1), D1
(3)aₙ = (aₙ − L) + L, so |aₙ| ≤ |aₙ − L| + |L|.A9, A2
(4)If n ≥ N, then |aₙ − L| + |L| < 1 + |L|.(2), A8
(5)Therefore, if n ≥ N, then |aₙ| < 1 + |L|.(3), (4), A6
(6)Let F = { |a₁|, |a₂|, …, |a_N|, 1 + |L| }. F is a finite set and is nonempty.convention
(7)F has a maximum M.(6), A1
(8)If n < N, then |aₙ| is an element of F, so |aₙ| ≤ M.(6), (7)
(9)1 + |L| is also an element of F, so 1 + |L| ≤ M, and if n ≥ N, then |aₙ| < 1 + |L| ≤ M.(5), (6), (7), A6
(10)Every natural number n satisfies n < N or n ≥ N, so |aₙ| ≤ M for every n.(8), (9), A11
(11)Therefore (aₙ) is bounded.(10), D2
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Theorem T2
A bounded sequence does not necessarily converge.
Proof.
(1)Let aₙ = (−1)ⁿ.convention
(2)If n is even, aₙ = 1; if odd, aₙ = −1.(1), A13
(3)|1| = 1 and |−1| = |1| = 1, so |aₙ| ≤ 1 for every n.(2), A5, A4, A3
(4)Therefore (aₙ) is bounded.(3), D2
(5)Suppose lim aₙ = L.assumption
(6)For ε = 1 there is a natural number N such that |aₙ − L| < 1 whenever n ≥ N.(5), A5, D1
(7)2N ≥ N and 2N + 1 ≥ N.A12
(8)At n = 2N, aₙ = 1, so |1 − L| < 1.(2), (6), (7)
(9)At n = 2N + 1, aₙ = −1, so |−1 − L| < 1.(2), (6), (7)
(10)|−1 − L| = |−(1 + L)| = |1 + L| = |L − (−1)|A3, A14
(11)(1 − L) + (L − (−1)) = 1 − (−1) = 1 + 1 = 2A10, A14
(12)|2| ≤ |1 − L| + |L − (−1)|(11), A2
(13)|1 − L| + |L − (−1)| < 1 + |L − (−1)| < 1 + 1 = 2(8), (9), (10), A8, A6, A14
(14)0 < 2, so |2| = 2, giving 2 < 2, a contradiction.(12), (13), A5, A4, A6
(15)Therefore ((−1)ⁿ) does not converge.(14), A7
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