LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01

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A geometric bound for tails of the exponential series

Statement

If $x\in\mathbb R$ , $N\in\mathbb N$ , and $\iota(N+2)\ge2|x|$ , then $\sum_{k=N+1}^{\infty}\frac{|x|^k}{\iota(k!)} \le \frac{2|x|^{N+1}}{\iota((N+1)!)}.$

Facts & Assumptions

Given: $x,N$ with the stated inequality.

[L1]

Factorials satisfy the recurrence; the canonical embedding preserves products and order and is strictly increasing on naturals (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $\mathbb{N}$ , The canonical natural $\iota(n) = n \cdot 1_F$ of a field, Canonical naturals are positive and strictly increasing).

[L2]

A geometric tail of ratio $1/2$ sums to $2$ (For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ the series diverges).

Proof

technique · direct

1.1

For $k\ge N+1$ , strict increase gives $\iota(k+1)\ge\iota(N+2)\ge2|x|$ , and the factorial recurrence gives that the ratio of consecutive absolute terms is $|x|/\iota(k+1)\le1/2$ .

givenL1

2.1

Thus the $j$ -th term after $N+1$ is at most the first tail term times $2^{-j}$ .

step 1.1induction

3.1

Sum the geometric majorant using [L2] to obtain the displayed bound.

step 2.1L2given∎

Depends on

Used by

The elementary numerical bound 2<e<3 Corollary
The number e is irrational Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 91 results over 23 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)
J. Lebl, Basic Analysis, Analytic Functions (standard reference, not scraped)
MIT Proofs in Analysis and Probability, Lecture 2 notes (standard reference, not scraped)
LSU MATH 7230, Homework 1 (standard reference, not scraped)