Metamath Proof Explorer


Theorem isumnn0nn

Description: Sum from 0 to infinity in terms of sum from 1 to infinity. (Contributed by NM, 2-Jan-2006) (Revised by Mario Carneiro, 24-Apr-2014)

Ref Expression
Hypotheses isumnn0nn.1 ( 𝑘 = 0 → 𝐴 = 𝐵 )
isumnn0nn.2 ( ( 𝜑𝑘 ∈ ℕ0 ) → ( 𝐹𝑘 ) = 𝐴 )
isumnn0nn.3 ( ( 𝜑𝑘 ∈ ℕ0 ) → 𝐴 ∈ ℂ )
isumnn0nn.4 ( 𝜑 → seq 0 ( + , 𝐹 ) ∈ dom ⇝ )
Assertion isumnn0nn ( 𝜑 → Σ 𝑘 ∈ ℕ0 𝐴 = ( 𝐵 + Σ 𝑘 ∈ ℕ 𝐴 ) )

Proof

Step Hyp Ref Expression
1 isumnn0nn.1 ( 𝑘 = 0 → 𝐴 = 𝐵 )
2 isumnn0nn.2 ( ( 𝜑𝑘 ∈ ℕ0 ) → ( 𝐹𝑘 ) = 𝐴 )
3 isumnn0nn.3 ( ( 𝜑𝑘 ∈ ℕ0 ) → 𝐴 ∈ ℂ )
4 isumnn0nn.4 ( 𝜑 → seq 0 ( + , 𝐹 ) ∈ dom ⇝ )
5 nn0uz 0 = ( ℤ ‘ 0 )
6 0zd ( 𝜑 → 0 ∈ ℤ )
7 5 6 2 3 4 isum1p ( 𝜑 → Σ 𝑘 ∈ ℕ0 𝐴 = ( ( 𝐹 ‘ 0 ) + Σ 𝑘 ∈ ( ℤ ‘ ( 0 + 1 ) ) 𝐴 ) )
8 fveq2 ( 𝑘 = 0 → ( 𝐹𝑘 ) = ( 𝐹 ‘ 0 ) )
9 8 1 eqeq12d ( 𝑘 = 0 → ( ( 𝐹𝑘 ) = 𝐴 ↔ ( 𝐹 ‘ 0 ) = 𝐵 ) )
10 2 ralrimiva ( 𝜑 → ∀ 𝑘 ∈ ℕ0 ( 𝐹𝑘 ) = 𝐴 )
11 0nn0 0 ∈ ℕ0
12 11 a1i ( 𝜑 → 0 ∈ ℕ0 )
13 9 10 12 rspcdva ( 𝜑 → ( 𝐹 ‘ 0 ) = 𝐵 )
14 0p1e1 ( 0 + 1 ) = 1
15 14 fveq2i ( ℤ ‘ ( 0 + 1 ) ) = ( ℤ ‘ 1 )
16 nnuz ℕ = ( ℤ ‘ 1 )
17 15 16 eqtr4i ( ℤ ‘ ( 0 + 1 ) ) = ℕ
18 17 sumeq1i Σ 𝑘 ∈ ( ℤ ‘ ( 0 + 1 ) ) 𝐴 = Σ 𝑘 ∈ ℕ 𝐴
19 18 a1i ( 𝜑 → Σ 𝑘 ∈ ( ℤ ‘ ( 0 + 1 ) ) 𝐴 = Σ 𝑘 ∈ ℕ 𝐴 )
20 13 19 oveq12d ( 𝜑 → ( ( 𝐹 ‘ 0 ) + Σ 𝑘 ∈ ( ℤ ‘ ( 0 + 1 ) ) 𝐴 ) = ( 𝐵 + Σ 𝑘 ∈ ℕ 𝐴 ) )
21 7 20 eqtrd ( 𝜑 → Σ 𝑘 ∈ ℕ0 𝐴 = ( 𝐵 + Σ 𝑘 ∈ ℕ 𝐴 ) )