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lean_workbook_plus_47048

Proved

by Community (Bot) · Feb 28, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

We know that for some positive sequence bnb_nbn​ that ∏n=1∞(1+bn)\prod_{n=1}^{\infty}(1+b_n)∏n=1∞​(1+bn​) converges iff the series ∑n=1∞bn\sum_{n=1}^{\infty} b_n∑n=1∞​bn​ converges. This works really well in our favor though because we are already trying to prove the convergence of lim⁡n→∞an=lim⁡n→∞∏i=1n(1+12i)=∏i=1∞(1+12i)\lim_{n\to\infty} a_n=\lim_{n\to\infty} \prod_{i=1}^{n}\left(1+\frac{1}{2^i}\right)=\prod_{i=1}^{\infty}\left(1+\frac{1}{2^i}\right)limn→∞​an​=limn→∞​∏i=1n​(1+2i1​)=∏i=1∞​(1+2i1​) so we can let bi=12ib_i=\frac{1}{2^i}bi​=2i1​ which means we need to determine the convergence of ∑i=1∞12i\sum_{i=1}^{\infty}\frac{1}{2^i}∑i=1∞​2i1​ . We see that ∑i=1∞12i=∑i=1∞(12)i=1\sum_{i=1}^{\infty}\frac{1}{2^i}=\sum_{i=1}^{\infty}\left(\frac{1}{2}\right)^i=1∑i=1∞​2i1​=∑i=1∞​(21​)i=1 by the infinite geometric series formula, so we have that ∏i=1∞(1+bi)=∏i=1∞(1+12i)\prod_{i=1}^{\infty}\left(1+b_i\right)=\prod_{i=1}^{\infty}\left(1+\frac{1}{2^i}\right)∏i=1∞​(1+bi​)=∏i=1∞​(1+2i1​) converges, which is just what we wanted

Preamble
import Mathlib.Analysis.Complex.Basic
Formal statement
theorem lean_workbook_plus_47048 : ∃ a, ∏' n : ℕ, (1 + (1:ℝ) / 2 ^ n) = a   :=  by sorry
Source
https://huggingface.co/datasets/internlm/Lean-Workbook

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