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Erdős Problem 68

Reference: erdosproblems.com/68

namespace Erdos68

Is $$\sum_{n=2}^\infty \frac{1}{n!-1}$$ irrational?

@[category research open, AMS 11] theorem erdos_68 : answer(sorry) Irrational (∑' n : , 1 / ((n + 2).factorial - 1 : )) := True Irrational (∑' (n : ), 1 / ((n + 2).factorial - 1)) All goals completed! 🐙

$$\sum_{n=2}^\infty \frac{1}{n!-1} = \sum_{n=2}^\infty \sum_{k=1}^\infty \frac{1}{(n!)^k}$$

f: := fun n k 1 / (n + 2).factorial ^ (k + 1)n:r: := (↑(n + 2).factorial)⁻¹hr_def:r = (↑(n + 2).factorial)⁻¹hr_nonneg:0 rhr_lt_one:r < 1hgeom:HasSum (fun n r ^ n) (1 - r)⁻¹hshift:HasSum (fun i r * r ^ i) (r * (1 - r)⁻¹)hf_eq: (k : ), f n k = r * r ^ k∑' (k : ), f n k = 1 / ((n + 2).factorial - 1) -- Evaluate ∑' k, f n k = r * (1 - r)⁻¹ = 1 / ((n + 2)! - 1). exact ((hshift.congr_fun hf_eq).tsum_eq.trans (f: := fun n k 1 / (n + 2).factorial ^ (k + 1)n:r: := (↑(n + 2).factorial)⁻¹hr_def:r = (↑(n + 2).factorial)⁻¹hr_nonneg:0 rhr_lt_one:r < 1hgeom:HasSum (fun n r ^ n) (1 - r)⁻¹hshift:HasSum (fun i r * r ^ i) (r * (1 - r)⁻¹)hf_eq: (k : ), f n k = r * r ^ kr * (1 - r)⁻¹ = 1 / ((n + 2).factorial - 1) f: := fun n k 1 / (n + 2).factorial ^ (k + 1)n:r: := (↑(n + 2).factorial)⁻¹hr_def:r = (↑(n + 2).factorial)⁻¹hr_nonneg:0 rhr_lt_one:r < 1hgeom:HasSum (fun n r ^ n) (1 - r)⁻¹hshift:HasSum (fun i r * r ^ i) (r * (1 - r)⁻¹)hf_eq: (k : ), f n k = r * r ^ k(↑(n + 2).factorial)⁻¹ * (1 - (↑(n + 2).factorial)⁻¹)⁻¹ = 1 / ((n + 2).factorial - 1); All goals completed! 🐙))end Erdos68