# Infinite series : convergence.

• Jul 11th 2008, 01:32 PM
arbolis
Infinite series : convergence.
$\displaystyle \sum_{n=1}^{+\infty} \frac{2^nn!}{n^n}$. I must show whether it converges or not. Using my intuition I think it converges. I tried to use the root test, the only test I think that works.
So I'm stuck when I must find the limit when $\displaystyle n$ tends to $\displaystyle +\infty$ of $\displaystyle \frac{2\cdot n!^{\frac{1}{n}}}{n}$. Does the limit of $\displaystyle n!^\frac{1}{n}=\frac{1}{e}$, when $\displaystyle n$ tends to $\displaystyle +\infty$? If yes, how to prove it using maths of calculus II?
• Jul 11th 2008, 01:37 PM
Mathstud28
Quote:

Originally Posted by arbolis
$\displaystyle \sum_{n=1}^{+\infty} \frac{2^nn!}{n^n}$. I must show whether it converges or not. Using my intuition I think it converges. I tried to use the root test, the only test I think that works.
So I'm stuck when I must find the limit when $\displaystyle n$ tends to $\displaystyle +\infty$ of $\displaystyle \frac{2\cdot n!^{\frac{1}{n}}}{n}$. Does the limit of $\displaystyle n!^\frac{1}{n}=\frac{1}{e}$, when $\displaystyle n$ tends to $\displaystyle +\infty$? If yes, how to prove it using maths of calculus II?

Root test.

And just consider that all this is limit is is

$\displaystyle \lim_{n\to\infty}(2^n)^{\frac{1}{n}}\cdot\lim_{n\t o\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}= 2\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{\fr ac{1}{n}}$

The last limit we have gone over.
• Jul 11th 2008, 01:40 PM
Plato
$\displaystyle \left( {\sqrt[n]{{\frac{{2^n n!}}{{n^n }}}}} \right) \to \frac{2}{e}$
• Jul 11th 2008, 01:44 PM
Mathstud28
Quote:

Originally Posted by arbolis
$\displaystyle \sum_{n=1}^{+\infty} \frac{2^nn!}{n^n}$. I must show whether it converges or not. Using my intuition I think it converges. I tried to use the root test, the only test I think that works.
So I'm stuck when I must find the limit when $\displaystyle n$ tends to $\displaystyle +\infty$ of $\displaystyle \frac{2\cdot n!^{\frac{1}{n}}}{n}$. Does the limit of $\displaystyle n!^\frac{1}{n}=\frac{1}{e}$, when $\displaystyle n$ tends to $\displaystyle +\infty$? If yes, how to prove it using maths of calculus II?

you could just say that $\displaystyle \frac{2^nn!}{n^n}\sim\left(\frac{2}{e}\right)^n$

By stirlings approximation
• Jul 11th 2008, 01:50 PM
Jhevon
the ratio test also works nicely here. you end up with the limit being $\displaystyle \frac 2e$ as well, which is less than 1, of course. it is easier to see the limit using this way, in my humble opinion
• Jul 11th 2008, 02:04 PM
arbolis
Thanks to all!! This means that $\displaystyle \sum_{n=1}^{+\infty} \frac{3^nn!}{n^n}$ diverges. (I changed the 2 for a 3). As I have to determine it as well, it's already done!
• Jul 11th 2008, 02:09 PM
Jhevon
Quote:

Originally Posted by arbolis
Thanks to all!! This means that $\displaystyle \sum_{n=1}^{+\infty} \frac{3^nn!}{n^n}$ diverges. (I changed the 2 for a 3). As I have to determine it as well, it's already done!

yes