Metamath Proof Explorer


Theorem clim2ser

Description: The limit of an infinite series with an initial segment removed. (Contributed by Paul Chapman, 9-Feb-2008) (Revised by Mario Carneiro, 1-Feb-2014)

Ref Expression
Hypotheses clim2ser.1 ⊢ Z = ℤ ≥ M
clim2ser.2 ⊢ φ → N ∈ Z
clim2ser.4 ⊢ φ ∧ k ∈ Z → F ⁡ k ∈ ℂ
clim2ser.5 ⊢ φ → seq M + F ⇝ A
Assertion clim2ser ⊢ φ → seq N + 1 + F ⇝ A − seq M + F ⁡ N

Proof

Step Hyp Ref Expression
1 clim2ser.1 ⊢ Z = ℤ ≥ M
2 clim2ser.2 ⊢ φ → N ∈ Z
3 clim2ser.4 ⊢ φ ∧ k ∈ Z → F ⁡ k ∈ ℂ
4 clim2ser.5 ⊢ φ → seq M + F ⇝ A
5 eqid ⊢ ℤ ≥ N + 1 = ℤ ≥ N + 1
6 2 1 eleqtrdi ⊢ φ → N ∈ ℤ ≥ M
7 peano2uz ⊢ N ∈ ℤ ≥ M → N + 1 ∈ ℤ ≥ M
8 6 7 syl ⊢ φ → N + 1 ∈ ℤ ≥ M
9 eluzelz ⊢ N + 1 ∈ ℤ ≥ M → N + 1 ∈ ℤ
10 8 9 syl ⊢ φ → N + 1 ∈ ℤ
11 eluzel2 ⊢ N ∈ ℤ ≥ M → M ∈ ℤ
12 6 11 syl ⊢ φ → M ∈ ℤ
13 1 12 3 serf ⊢ φ → seq M + F : Z ⟶ ℂ
14 13 2 ffvelcdmd ⊢ φ → seq M + F ⁡ N ∈ ℂ
15 seqex ⊢ seq N + 1 + F ∈ V
16 15 a1i ⊢ φ → seq N + 1 + F ∈ V
17 13 adantr ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq M + F : Z ⟶ ℂ
18 8 1 eleqtrrdi ⊢ φ → N + 1 ∈ Z
19 1 uztrn2 ⊢ N + 1 ∈ Z ∧ j ∈ ℤ ≥ N + 1 → j ∈ Z
20 18 19 sylan ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → j ∈ Z
21 17 20 ffvelcdmd ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq M + F ⁡ j ∈ ℂ
22 addcl ⊢ k ∈ ℂ ∧ x ∈ ℂ → k + x ∈ ℂ
23 22 adantl ⊢ φ ∧ j ∈ ℤ ≥ N + 1 ∧ k ∈ ℂ ∧ x ∈ ℂ → k + x ∈ ℂ
24 addass ⊢ k ∈ ℂ ∧ x ∈ ℂ ∧ y ∈ ℂ → k + x + y = k + x + y
25 24 adantl ⊢ φ ∧ j ∈ ℤ ≥ N + 1 ∧ k ∈ ℂ ∧ x ∈ ℂ ∧ y ∈ ℂ → k + x + y = k + x + y
26 simpr ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → j ∈ ℤ ≥ N + 1
27 6 adantr ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → N ∈ ℤ ≥ M
28 elfzuz ⊢ k ∈ M … j → k ∈ ℤ ≥ M
29 28 1 eleqtrrdi ⊢ k ∈ M … j → k ∈ Z
30 29 3 sylan2 ⊢ φ ∧ k ∈ M … j → F ⁡ k ∈ ℂ
31 30 adantlr ⊢ φ ∧ j ∈ ℤ ≥ N + 1 ∧ k ∈ M … j → F ⁡ k ∈ ℂ
32 23 25 26 27 31 seqsplit ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq M + F ⁡ j = seq M + F ⁡ N + seq N + 1 + F ⁡ j
33 32 oveq1d ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq M + F ⁡ j − seq M + F ⁡ N = seq M + F ⁡ N + seq N + 1 + F ⁡ j - seq M + F ⁡ N
34 14 adantr ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq M + F ⁡ N ∈ ℂ
35 1 uztrn2 ⊢ N + 1 ∈ Z ∧ k ∈ ℤ ≥ N + 1 → k ∈ Z
36 18 35 sylan ⊢ φ ∧ k ∈ ℤ ≥ N + 1 → k ∈ Z
37 36 3 syldan ⊢ φ ∧ k ∈ ℤ ≥ N + 1 → F ⁡ k ∈ ℂ
38 5 10 37 serf ⊢ φ → seq N + 1 + F : ℤ ≥ N + 1 ⟶ ℂ
39 38 ffvelcdmda ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq N + 1 + F ⁡ j ∈ ℂ
40 34 39 pncan2d ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq M + F ⁡ N + seq N + 1 + F ⁡ j - seq M + F ⁡ N = seq N + 1 + F ⁡ j
41 33 40 eqtr2d ⊢ φ ∧ j ∈ ℤ ≥ N + 1 → seq N + 1 + F ⁡ j = seq M + F ⁡ j − seq M + F ⁡ N
42 5 10 4 14 16 21 41 climsubc1 ⊢ φ → seq N + 1 + F ⇝ A − seq M + F ⁡ N