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Metamath Proof Explorer

Theorem geoihalfsum

Description: Prove that the infinite geometric series of 1/2, 1/2 + 1/4 + 1/8 + ... = 1. Uses geoisum1 . This is a representation of .111... in binary with an infinite number of 1's. Theorem 0.999... proves a similar claim for .999... in base 10. (Contributed by David A. Wheeler, 4-Jan-2017) (Proof shortened by AV, 9-Jul-2022)

Ref Expression
Assertion geoihalfsum Σ 𝑘 ∈ ℕ ( 1 / ( 2 ↑ 𝑘 ) ) = 1

Proof

Step Hyp Ref Expression
1 2cn ⊢ 2 ∈ ℂ
2 1 a1i ⊢ ( 𝑘 ∈ ℕ → 2 ∈ ℂ )
3 2ne0 ⊢ 2 ≠ 0
4 3 a1i ⊢ ( 𝑘 ∈ ℕ → 2 ≠ 0 )
5 nnz ⊢ ( 𝑘 ∈ ℕ → 𝑘 ∈ ℤ )
6 2 4 5 exprecd ⊢ ( 𝑘 ∈ ℕ → ( ( 1 / 2 ) ↑ 𝑘 ) = ( 1 / ( 2 ↑ 𝑘 ) ) )
7 6 sumeq2i ⊢ Σ 𝑘 ∈ ℕ ( ( 1 / 2 ) ↑ 𝑘 ) = Σ 𝑘 ∈ ℕ ( 1 / ( 2 ↑ 𝑘 ) )
8 halfcn ⊢ ( 1 / 2 ) ∈ ℂ
9 halfre ⊢ ( 1 / 2 ) ∈ ℝ
10 halfge0 ⊢ 0 ≤ ( 1 / 2 )
11 absid ⊢ ( ( ( 1 / 2 ) ∈ ℝ ∧ 0 ≤ ( 1 / 2 ) ) → ( abs ‘ ( 1 / 2 ) ) = ( 1 / 2 ) )
12 9 10 11 mp2an ⊢ ( abs ‘ ( 1 / 2 ) ) = ( 1 / 2 )
13 halflt1 ⊢ ( 1 / 2 ) < 1
14 12 13 eqbrtri ⊢ ( abs ‘ ( 1 / 2 ) ) < 1
15 geoisum1 ⊢ ( ( ( 1 / 2 ) ∈ ℂ ∧ ( abs ‘ ( 1 / 2 ) ) < 1 ) → Σ 𝑘 ∈ ℕ ( ( 1 / 2 ) ↑ 𝑘 ) = ( ( 1 / 2 ) / ( 1 − ( 1 / 2 ) ) ) )
16 8 14 15 mp2an ⊢ Σ 𝑘 ∈ ℕ ( ( 1 / 2 ) ↑ 𝑘 ) = ( ( 1 / 2 ) / ( 1 − ( 1 / 2 ) ) )
17 1mhlfehlf ⊢ ( 1 − ( 1 / 2 ) ) = ( 1 / 2 )
18 17 oveq2i ⊢ ( ( 1 / 2 ) / ( 1 − ( 1 / 2 ) ) ) = ( ( 1 / 2 ) / ( 1 / 2 ) )
19 ax-1cn ⊢ 1 ∈ ℂ
20 ax-1ne0 ⊢ 1 ≠ 0
21 19 1 20 3 divne0i ⊢ ( 1 / 2 ) ≠ 0
22 8 21 dividi ⊢ ( ( 1 / 2 ) / ( 1 / 2 ) ) = 1
23 16 18 22 3eqtri ⊢ Σ 𝑘 ∈ ℕ ( ( 1 / 2 ) ↑ 𝑘 ) = 1
24 7 23 eqtr3i ⊢ Σ 𝑘 ∈ ℕ ( 1 / ( 2 ↑ 𝑘 ) ) = 1