# Thread: limit equal to 0?

1. ## limit equal to 0?

Hello everyone;

This might be a silly question, but what is the limit of $\displaystyle \frac{\pi\sqrt[n]{n!}}{n}$ when $\displaystyle n$ tends to infinity? To my mind, it would be 0. What do you think?

2. Can you use the inequality $\displaystyle \left( {n - 1} \right)! \le n^n e^{ - n} e \le n!$ to prove the following?
$\displaystyle \lim _{n \to \infty } \left( {\frac{{\sqrt[n]{{n!}}}}{{n }}} \right) = \frac{1}{e}$

3. Actually, this limit is not 0.

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

So, with the Pi, your limit is $\displaystyle \frac{\pi}{e}$

EDIT: Oops, Plato beat me.

4. Originally Posted by Plato
Can you use the inequality $\displaystyle \left( {n - 1} \right)! \le n^n e^{ - n} e \le n!$ to prove the following?
$\displaystyle \lim _{n \to \infty } \left( {\frac{{\sqrt[n]{{n!}}}}{{n }}} \right) = \frac{1}{e}$
where did that inequality come from?

5. Originally Posted by galactus
Actually, this limit is not 0.

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

So, with the Pi, your limit is $\displaystyle \frac{\pi}{e}$

EDIT: Oops, Plato beat me.
O wait, i did this problem already. i should have known this (TPH would be so disappointed in me)

6. Originally Posted by Jhevon
where did that inequality come from?
Here is a slightly different version.

7. It's been bugging me for quite long, I wanted to determine if a series $\displaystyle \sum_{n=1}^{\infty}\left(\frac{\pi}{n}\right)^nn!$ is convergent or divergent.

By d'Alembert's criterion:

$\displaystyle a_{n}=\left(\frac{\pi}{n}\right)^{n}(n)!$
$\displaystyle a_{n+1}=\left(\frac{\pi}{n+1}\right)^{n+1}(n+1)!$
$\displaystyle \lim_{n\rightarrow{\infty}}\left|\frac{a_{n+1}}{a_ n}\right|=\lim_{n\rightarrow{\infty}}\frac{\left(\ frac{\pi}{n+1}\right)^{n+1}(n+1)!}{\left(\frac{\pi }{n}\right)^nn!}=\lim_{n\rightarrow{\infty}}\frac{ \left(\frac{\pi}{n}\right)^n\left(\frac{\pi}{n}\ri ght)n!(n+1)}{\left(\frac{\pi}{n}\right)^nn!}=$
$\displaystyle =\lim_{n\rightarrow{\infty}}\left[\pi\left(\frac{n}{n+1}\right)^n\right]=\lim_{n\rightarrow{\infty}}\left[\pi\left(1+\frac{1}{n}\right)^n\right]^{-1}=\pi{e}^{-1}=\frac{\pi}{e}$

Then $\displaystyle \frac{\pi}{e}>1$, so $\displaystyle \sum_{n=1}^{\infty}\left(\frac{\pi}{n}\right)^nn!$ is divergent.

By Cauchy's criterion:

$\displaystyle \lim_{n\rightarrow{\infty}}\sqrt[n]{a_n}=\lim_{n\rightarrow{\infty}}\sqrt[n]{\left(\frac{\pi}{n}\right)^nn!}=\lim_{n\rightarro w{\infty}}\frac{\pi}{n}\sqrt[n]{n!}=\lim_{n\rightarrow{\infty}}\frac{\pi\sqrt[n]{n!}}{n}=?$

That's where I was left puzzled. Were the limit equal to zero, then by Cauchy's criterion the series would be convergent. Now that I know it's also $\displaystyle \frac{\pi}{e}$, I have no more doubts.

Thanks a lot for your replies.

8. Originally Posted by Jhevon
where did that inequality come from?
I haven't bothered to look it up, but it's likely a variation on Stirling's formula for n! (where n is very large.)

-Dan

9. What are $\displaystyle e$, $\displaystyle \forall$, ln?

10. Originally Posted by SengNee
What are $\displaystyle e$, $\displaystyle \forall$, ln?

"e" is "Napier's constant," the number 2.718281828459045235360287471352662497757... This number has a lot of special properties, which I'll let you look up.

$\displaystyle \forall$ is a Math hieroglyph for the words "for all." It is most commonly used in set theory.

"ln()" is a logarithmic function, for example: $\displaystyle f(x) = \log_e(x) = ln(x)$, where e is Napier's number above.

-Dan

11. You can use the simple fact. If $\displaystyle a_n$ is a sequence of non-zero terms such that $\displaystyle a_{n+1}/a_n \to L$ then $\displaystyle |a_n|^{1/n}\to L$ also.

(This was an excercise in section 12 # 14 in Ross book, if I remember correcly).

Thus, if you let $\displaystyle a_n = \frac{n!}{n^n}$ then $\displaystyle \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n!}{n^n} \to \frac{1}{e}$.

That means, $\displaystyle (a_n)^{1/n} = \frac{1}{n}(n!)^{1/n} \to \frac{1}{e}$.

12. Originally Posted by disclaimer
Hello everyone;

This might be a silly question, but what is the limit of $\displaystyle \frac{\pi\sqrt[n]{n!}}{n}$ when $\displaystyle n$ tends to infinity? To my mind, it would be 0. What do you think?

There is a nice trick with factorials, which is like killing a fly with a cannon, but it works well over here.

Sitriling approximation says you can replace $\displaystyle n!$ by $\displaystyle \sqrt{2\pi} \cdot \sqrt{n} \cdot n^n \cdot e^{-n}$.

That means,
$\displaystyle \frac{\pi (n!)^{1/n}}{n} \to \frac{\pi \left[ \sqrt{2\pi} \cdot \sqrt{n} \cdot n^n \cdot e^{-n} \right]^{1/n}}{n} \to \frac{\pi}{e}$.

13. Originally Posted by topsquark
"e" is "Napier's constant," the number 2.718281828459045235360287471352662497757... This number has a lot of special properties, which I'll let you look up.

$\displaystyle \forall$ is a Math hieroglyph for the words "for all." It is most commonly used in set theory.

"ln()" is a logarithmic function, for example: $\displaystyle f(x) = \log_e(x) = ln(x)$, where e is Napier's number above.

-Dan
Can you show me some properties of $\displaystyle e$ and how to use $\displaystyle \forall$ in set theory?

14. Originally Posted by SengNee
Can you sow me some properties of $\displaystyle e$ and how to use $\displaystyle \forall$ in set theory?
Here are some properties of $\displaystyle e$.
• $\displaystyle \lim ~ \left( 1 + \frac{1}{n} \right)^n = e$
• $\displaystyle 1+\frac{1}{1!}+\frac{1}{2!}+... = \sum_{n=0}^{\infty} \frac{1}{n!} = e$
• $\displaystyle \int_1^e \frac{dt}{t} = 1$ --> this is what I use as the definition of $\displaystyle e$.

15. Originally Posted by ThePerfectHacker
Here are some properties of $\displaystyle e$.
• $\displaystyle \lim ~ \left( 1 + \frac{1}{n} \right)^n = e$
• $\displaystyle 1+\frac{1}{1!}+\frac{1}{2!}+... = \sum_{n=0}^{\infty} \frac{1}{n!} = e$
• $\displaystyle \int_1^e \frac{dt}{t} = 1$ --> this is what I use as the definition of $\displaystyle e$.

$\displaystyle \sum_{n=0}^{\infty} \frac{1}{n!} =1+\frac{1}{1!}+\frac{1}{2!}+... = e$

or

$\displaystyle \sum_{n=0}^{\infty} \frac{1}{n!} =0+1+\frac{1}{1!}+\frac{1}{2!}+... = e$

???

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