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Elementary Proofs of Some Stirling Bounds

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Nader H. Bshouty, Vivian E. Bshouty-Hurani, George Haddad, Thomas Hashem, Fadi Khoury, Omar Sharafy

Functional Analysis · Submitted 20 Feb 2018

abstract compiled · 2026-09-23 · no language-model review

The authors' main claim In particular, we show that for all $n\ge 8$ $$\sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n}-\frac{1}{360n^3+103n}} \ge n!\ge \sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n}-\frac{1}{360n^3+102n}}$$ and for all $n\ge 3$ $$\sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n+\frac{2}{5n}-\frac{1.1}{10n^3}}} \ge n!\ge \sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n+\frac{2}{5n}-\frac{0.9}{10n^3}}}.$$

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arXiv ID1802.07046 (version 2)
Pith IDK3KNOP4V, Pith's permanent record of this paper and its review
CitationFadi Khoury, George Haddad, Nader H. Bshouty, Omar Sharafy, Thomas Hashem, Vivian E. Bshouty-Hurani. "Elementary Proofs of Some Stirling Bounds." arXiv:1802.07046 (2018). https://pith.science/paper/K3KNOP4V
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Compiled by classifier-1.13.0 via TypeSafe API, question bank abstract-p1-20260923, 2026-09-23: 1 request, 3,292 input tokens.

What the abstract itself states, as probabilities: claims first 0.05 · states limitation 0.13 · human participants 0.01 · controlled comparison 0.03 · randomized assignment 0.01 · states quantitative result 0.98 · compares with prior results 0.87 · mentions code or data availability 0.02.

The abstract, sentence by sentence

arXiv listing abstract v2. Results and the main claim are underlined; .

We give elementary proofs of several Stirling's precise bounds. We first improve all the precise bounds from the literature and give new precise bounds. In particular, we show that for all $n\ge 8$ $$\sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n}-\frac{1}{360n^3+103n}} \ge n!\ge \sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n}-\frac{1}{360n^3+102n}}$$ and for all $n\ge 3$ $$\sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n+\frac{2}{5n}-\frac{1.1}{10n^3}}} \ge n!\ge \sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n+\frac{2}{5n}-\frac{0.9}{10n^3}}}.$$

background contribution method result interpretation main claim

What the classifier reports for each sentence. The number is the probability it gave the label.
#sentencerolethe authors' own workstates the main conclusionstrength
s0 We give elementary proofs of several Stirling's precise bounds. contribution 0.99 0.96 0.56 asserted 1.00
s1 We first improve all the precise bounds from the literature and give new precise bounds. contribution 0.99 0.95 0.23 asserted 1.00
s2 In particular, we show that for all $n\ge 8$ $$\sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n}-\frac{1}{360n^3+103n}} \ge n!\ge \sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n}-\frac{1}{360n^3+102n}}$$ and for all $n\ge 3$ $$\sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n+\frac{2}{5n}-\frac{1.1}{10n^3}}} \ge n!\ge \sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n+\frac{2}{5n}-\frac{0.9}{10n^3}}}.$$ result 0.97 0.96 0.76 asserted 1.00

What kind of paper the classifier reports this is: theoretical result 1.00 · evidence from proof or derivation 1.00.

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  • full text
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  • references and citing sentences
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  • T body sentences
  • C citations
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Every quoted sentence on this record is the authors'. Every label is from Pith's closed vocabulary, with the probability shown beside it. Every number in the measurements table is the authors', checked by code. Record as JSON

The paper
Original abstract

We give elementary proofs of several Stirling's precise bounds. We first improve all the precise bounds from the literature and give new precise bounds. In particular, we show that for all $n\ge 8$ $$\sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n}-\frac{1}{360n^3+103n}} \ge n!\ge \sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n}-\frac{1}{360n^3+102n}}$$ and for all $n\ge 3$ $$\sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n+\frac{2}{5n}-\frac{1.1}{10n^3}}} \ge n!\ge \sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{1}{12n+\frac{2}{5n}-\frac{0.9}{10n^3}}}.$$

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