Definition D1
The value of the infinite geometric series a + ar + ar² + … is defined as the limit of the partial-sum sequence (Sₙ).
Proof.
(1)Set S₁ = a, Sₙ₊₁ = Sₙ + a rⁿ. Sₙ is the sum of the first n terms.convention
(2)The phrase “adding infinitely” has no meaning outside this convention. Whether a value exists depends on whether the limit exists.convention
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Definition D2
lim xₙ = L means the following. For every ε > 0 there is a natural number N such that |xₙ − L| < ε for every n ≥ N.
Proof.
If such an L exists, the sequence is said to converge; if no such L exists, it is said to diverge.convention
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Lemma L1
If |r| < 1, then lim rⁿ = 0
Proof.
(1)Set s = |r|. Then 0 ≤ s.convention, A10
(2)First consider the case r = 0. For every n, rⁿ = 0.A18, A13
(3)For every ε > 0, |rⁿ − 0| = |0| = 0 < ε, so lim rⁿ = 0.(2), A19, A11, D2
(4)Now suppose r ≠ 0. Then 0 < s < 1.assumption, (1), A11
(5)Setting t = 1/s gives s·t = 1 and 1 < t.(4), A9, A7
(6)There is an h > 0 with t = 1 + h.(5), A6
(7)Suppose ε with 0 < ε is given.assumption
(8)0 < hε.(6), (7), A3, A18
(9)There is a natural number N with N(hε) > 1.(8), A1
(10)If n ≥ N, then N(hε) ≤ n(hε), so 1 < n(hε).(9), A3, E, A4
(11)n(hε) = (nh)εA21
(12)(1+h)ⁿ ≥ 1 + nh and nh < 1 + nh, so nh < tⁿ.(6), A2, A5, A4
(13)sⁿtⁿ = (st)ⁿ = 1ⁿ = 1(5), A20
(14)0 < sⁿ(4), A3, A18, A13
(15)sⁿ(nh) < sⁿtⁿ = 1(12), (13), (14), A3, A22
(16)sⁿ(nh) < 1 < (nh)ε, so sⁿ(nh) < (nh)ε.(10), (11), (15), A4
(17)0 < 1 ≤ n and 0 < h, so 0 < nh.(6), A5, A4, A3, A18
(18)Therefore sⁿ < ε.(16), (17), A3, A22
(19)|rⁿ − 0| = |rⁿ| = |r|ⁿ = sⁿ < ε(18), A19, A12, A13
(20)For every ε > 0 such an N exists, so lim rⁿ = 0.(7), (9), (19), D2
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Theorem T1
If |r| < 1, the infinite geometric series converges and its value is a/(1 − r).
Proof.
(1)r ≤ |r| < 1, so r < 1. If r = 1, then 1 < 1, a contradiction, so r ≠ 1.assumption, A10, A4, A8, E
(2)Sₙ = a(1−rⁿ)/(1−r)(1), A14
(3)Setting C = a/(1−r) gives Sₙ = C(1 − rⁿ).(2), A23, convention
(4)C(1−rⁿ) = C×1 − C rⁿ = C − C rⁿA24, A25
(5)lim rⁿ = 0L1
(6)lim C rⁿ = C · lim rⁿ = C × 0 = 0(5), A15, A18
(7)lim Sₙ = lim(C − C rⁿ) = C − lim C rⁿ = C − 0 = C(3), (4), (6), A16, A19
(8)Therefore the series converges and its value is a/(1 − r).(7), D1
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Corollary C
If a ≠ 0 and 1 ≤ |r|, then (Sₙ) diverges.
Proof.
(1)Sₙ₊₁ − Sₙ = a rⁿD1, A26
(2)|a rⁿ| = |a| |r|ⁿA12, A13
(3)1 ≤ |r|, so 1 ≤ |r|ⁿ.assumption, A13, A3, E
(4)a ≠ 0, so 0 < |a|.assumption, A11, A10
(5)|a| = |a|×1 ≤ |a| |r|ⁿ(3), (4), A25, A3, E
(6)Suppose (Sₙ) converges to some L.assumption
(7)Then Sₙ₊₁ − Sₙ → 0.(6), A17
(8)Taking ε = |a| in D2, from some N onward |(Sₙ₊₁ − Sₙ) − 0| < |a|.(4), (7), D2
(9)But |(Sₙ₊₁ − Sₙ) − 0| = |a rⁿ| and |a| ≤ |a rⁿ|.(1), (2), (5), A19
(10)Then |a| < |a|, a contradiction. Therefore (Sₙ) does not converge.(8), (9), A4, A8
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