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?
Thanks in advance.
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.
"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
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, $\displaystyle (a_n)^{1/n} = \frac{1}{n}(n!)^{1/n} \to \frac{1}{e}$.
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,
$\displaystyle \frac{\pi (n!)^{1/n}}{n} \to \frac{\pi \left[ \sqrt{2\pi} \cdot \sqrt{n} \cdot n^n \cdot e^{-n} \right]^{1/n}}{n} \to \frac{\pi}{e}$.
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 $\displaystyle e$.