Proof That $\lim_{n \rightarrow \infty} \frac{n}{(n!)^\frac{1}{n}} = e$

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Discussion Overview

The discussion centers on the limit $\lim_{n \rightarrow \infty} \frac{n}{(n!)^\frac{1}{n}}$ and whether it equals $e$. Participants explore various approaches to evaluate this limit, including the use of Stirling's formula and numerical calculations.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant questions whether the limit equals $e$ or if it is divergent.
  • Another participant suggests using Stirling's formula for $n!$ and mentions applying L'Hôpital's rule.
  • A different participant notes that Stirling's formula seems to imply the limit could be $e/\sqrt{2\pi}$, but this is not guaranteed.
  • One participant calculates an approximation of $e/\sqrt{2\pi}$ and compares it to numerical results from their calculations, which suggest a value around 2.67021.
  • Another participant provides a derivation using Stirling's approximation, concluding that the limit simplifies to $e$.

Areas of Agreement / Disagreement

Participants express differing views on the limit's value, with some suggesting it approaches $e$ and others proposing it might be $e/\sqrt{2\pi}$. The discussion remains unresolved as no consensus is reached.

Contextual Notes

Participants reference Stirling's approximation and numerical calculations, but there are unresolved assumptions regarding the application of these methods and the accuracy of the approximations used.

JonF
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does...

[tex]\lim_{n \rightarrow \infty} \frac{n}{(n!)^\frac{1}{n}} = e[/tex]

If not, is it divergent?
 
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isnt there somethiong called stirlings formula for n! ?? Maybe you could use that and lhopital.
 
well i just looked up stirling and it seems to suggest at a quick calculation, not guaranteed, that this limit is e/sqrt(2pi)
 
e/(2pi)^(1/2) aprox= 1.0844

my calc can do the limit up to 200 and it equals about 2.67021... that's why i thought it may = e
 
With Stirlings approximation: [itex]N!\approx N^Ne^{-N}[/itex], you indeed get:

[tex]\frac{N}{(N^Ne^{-N})^{\frac{1}{N}}}=\frac{N}{Ne^{-1}}=e[/tex]
 

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