A Beginner's Guide to the Squeeze Theorem

[karush]

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

 

[evinda]

Re: Squeeze theorem

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 $x$ ?

 

[karush]

Re: Squeeze theorem

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

 

[evinda]

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?

 

[Evgeny.Makarov]

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.

 

[Evgeny.Makarov]

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 $n\to\infty$.

 

[karush]

Ok, I see how this works now... was quite a bit more understandable than book examples.