A Beginner's Guide to the Squeeze Theorem

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Discussion Overview

The discussion revolves around the application of the squeeze theorem to evaluate the limit $\displaystyle \lim_{{n}\to{\infty}} \frac{n!}{{n}^{x}}=0$. Participants explore various approaches to demonstrate this limit, including comparisons of growth rates and convergence of series.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant suggests that the denominator $n^x$ increases faster than the numerator $n!$, leading to the conclusion that the limit approaches 0.
  • Another participant questions the information provided about the variable $x$ in the limit expression.
  • A participant corrects a typo regarding the limit expression, clarifying that $\displaystyle \lim_{{n}\to{\infty}} \frac{n!}{{n}^{n}}=0$ is the intended limit.
  • It is noted that if the series $\sum a_n$ converges, then $\lim_{n \to +\infty} a_n=0$, prompting a check on the convergence of $\sum_{n=1}^{\infty} \frac{n!}{n^n}$.
  • A participant provides a detailed breakdown of the limit using the squeeze theorem, demonstrating that the sequence is bounded and approaches 0.
  • Another participant offers a simpler upper bound for $\frac{n!}{n^n}$, showing that it is less than or equal to $\frac{1}{n}$, which also approaches 0 as $n$ increases.
  • A later reply expresses understanding of the concept, indicating that the explanation was clearer than examples found in textbooks.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the limit's evaluation method, as various approaches and interpretations are presented. Some participants focus on the squeeze theorem, while others consider convergence of series.

Contextual Notes

There are references to the relationship between limits and series convergence, but the discussion does not resolve the implications of these relationships. The variable $x$ remains unspecified, which may affect the limit's evaluation.

karush
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Use the squeeze theorem to show that
$\displaystyle
\lim_{{n}\to{\infty}} \frac{n!}{{n}^{x}}=0 \\
\text{have never used the squeeze theorem } \\
\text{but by observation the denominator is increasing faster}$
 
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Re: Squeeze theorem

karush said:
Use the squeeze theorem to show that
$\displaystyle
\lim_{{n}\to{\infty}} \frac{n!}{{n}^{x}}=0 \\
\text{have never used the squeeze theorem } \\
\text{but by observation the denominator is increasing faster}$

What information is given about $x$ ?
 
Re: Squeeze theorem

$\displaystyle
\lim_{{n}\to{\infty}} \frac{n!}{{n}^{n}}=0 \\
\text{sorry typo} $
 
Re: Squeeze theorem

It holds that if $\sum a_n$ converges, then $\lim_{n \to +\infty} a_n=0$.

Can you check if $\sum_{n=1}^{\infty} \frac{n!}{n^n}$ converges?
 
The problem asked to prove the limit using the squeeze theorem. In a calculus course limits of sequences (including the squeeze theorem) may come before series.

We have
\begin{multline}
0\le\frac{n!}{n^n}=
\frac{n\cdot(n-1)\cdot\ldots\cdot\left(\lfloor n/2\rfloor+1\right)}{n\cdot n\cdot\ldots\cdot n}\cdot\frac{\lfloor n/2\rfloor\cdot\left(\lfloor n/2\rfloor-1\right)\cdot\ldots\cdot1}{n\cdot n\cdot\ldots\cdot n}
\le 1\cdot\left(\frac{1}{2}\right)^{\lfloor n/2\rfloor}
\end{multline}
The sequence $(1/2)^{\lfloor n/2\rfloor}\to0$ as $n\to\infty$ (this can be proved by definition). Therefore, the original sequence is squeezed between two sequences whose limit is 0.
 
As Deveno pointed out, an even simpler upper bound is
\[
\dfrac{n!}{n^n} = \dfrac{1}{n}\cdot\dfrac{2}{n} \cdots\dfrac{n}{n} \leq \dfrac{1}{n}\cdot 1\cdots 1 = \dfrac{1}{n},
\]
and it is known that $1/n\to0$ as $n\to\infty$.
 
Ok, I see how this works now... was quite a bit more understandable than book examples.
 

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