Squeeze (Sandwich) Theorem question

[Cilius_Pippi]

lim n!/(n^n)
->inf


Solve using the Sandwich Theorem

 

[Cilius_Pippi]

I have to use the Sandwich rule to find the limit:

lim n->inf n!/(n^n)

I know it's 0.
I know how to use the squeeze theorem,
I just am not sure what the functions doing the squeezing should be, so that the theorem works. I am also told I have to use the Squeeze theorem.

Thanks for your help in advance,
Cilius

 

[stapel]

I have to use the Sandwich rule to find the limit:

lim n->inf n!/(n^n)

I am also told I have to use the Squeeze theorem.

What is the difference between the "Sandwich" and "Squeeze" Theorems in your book? (I've always thought they were the same thing.)

Thank you!

 

[Ishuda]

I have to use the Sandwich rule to find the limit:

lim n->inf n!/(n^n)

I know it's 0.
I know how to use the squeeze theorem,
I just am not sure what the functions doing the squeezing should be, so that the theorem works. I am also told I have to use the Squeeze theorem.

Thanks for your help in advance,
Cilius

What are your thoughts? What have you done so far? Please show us your work even if you feel that it is wrong so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting

You need to find two functions of n, call them f(n) and g(n), so that
\(\displaystyle \underset{n\, \to\, \infty}{lim}\, f(n)\, =\, \underset{n\, \to\, \infty}{lim}\, g(n)\)
and
\(\displaystyle f(n)\, \le\, \frac{n!}{n^n}\, \le\, g(n)\)

As a hint,
\(\displaystyle \frac{1}{3^2}\, =\, \frac{1\, 1\, 3}{3\, 3\, 3}\, \le\, \frac{1\, 2\, 3}{3\, 3\, 3}\, \le\, \frac{1\, 3\, 3}{3\, 3\, 3}\, =\, \frac{1}{3}\)

 

[stapel]

Note to original poster: When Ishuda states the following:

You need to find two functions of n, call them f(n) and g(n),....

...he isn't saying that you have to find the one "right" pair of functions, as though there were only two which might work. You can find any two that happen to be useful! You and a friend might pick entirely different pairs of functions, and yet both have "right" answers, because each of your pairs works.

So use your imagination! Don't feel constrained or limited. Think about stuff you've graphed, about functions that "feel" similar to you, etc. Have some fun with this!

 

[Ishuda]

Note to original poster: When Ishuda states the following: ...

Very true, and, in that spirit, for n>0
\(\displaystyle 0\, \le\, \frac{1}{n^5}\, \le\, \frac{1}{n^4}\, \le\,\, \frac{1}{n^3}\, \le\,\, \frac{1}{n^2}\)

 

[pka]

Very true, and, in that spirit, for n>0
\(\displaystyle 0\, \le\, \frac{1}{n^5}\, \le\, \frac{1}{n^4}\, \le\,\, \frac{1}{n^3}\, \le\,\, \frac{1}{n^2}\)

There is a rather nice way to do this by the sandwich method that results from this remarkable limit:
\(\displaystyle {\lim _{n \to \infty }}\left( {\dfrac{n}{{\sqrt[n]{{n!}}}}} \right) = e\). If you know the root test, the fact that the series \(\displaystyle \sum\limits_{n = 1}^\infty {\frac{{n!}}{{{n^n}}}} \) converges would prove the OP.

But to use the sandwich method:
​\(\displaystyle \begin{align*} \log\left[(n-1)!]\right]&\le\int_1^n {\log (x)dx} \le\log\left[n!\right]\\ \log\left[(n-1)!]\right]&\le n\log(n)-n+1 \le\log\left[n!\right]\\\dfrac{{{e^{n - 1}}}}{n} &\le \dfrac{{{n^n}}}{{n!}} \le {e^{n - 1}}\end{align*}\)

By using the inverse of the last line we see the limit.