Infinite series : convergence.

**arbolis** · July 11th 2008, 01:32 PM

$\sum_{n=1}^{+\infty} \frac{2^nn!}{n^n}$ . 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 $n$ tends to $+\infty$ of $\frac{2\cdot n!^{\frac{1}{n}}}{n}$ . Does the limit of $n!^\frac{1}{n}=\frac{1}{e}$ , when $n$ tends to $+\infty$ ? If yes, how to prove it using maths of calculus II?

**Mathstud28** · July 11th 2008, 01:37 PM

Originally Posted by arbolis

$\sum_{n=1}^{+\infty} \frac{2^nn!}{n^n}$ . 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 $n$ tends to $+\infty$ of $\frac{2\cdot n!^{\frac{1}{n}}}{n}$ . Does the limit of $n!^\frac{1}{n}=\frac{1}{e}$ , when $n$ tends to $+\infty$ ? If yes, how to prove it using maths of calculus II?

Root test.

And just consider that all this is limit is is

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

The last limit we have gone over.

**Plato** · July 11th 2008, 01:40 PM

$\left( {\sqrt[n]{{\frac{{2^n n!}}{{n^n }}}}} \right) \to \frac{2}{e}$

**Mathstud28** · July 11th 2008, 01:44 PM

Originally Posted by arbolis

$\sum_{n=1}^{+\infty} \frac{2^nn!}{n^n}$ . 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 $n$ tends to $+\infty$ of $\frac{2\cdot n!^{\frac{1}{n}}}{n}$ . Does the limit of $n!^\frac{1}{n}=\frac{1}{e}$ , when $n$ tends to $+\infty$ ? If yes, how to prove it using maths of calculus II?

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

By stirlings approximation

**Jhevon** · July 11th 2008, 01:50 PM

the ratio test also works nicely here. you end up with the limit being $\frac 2e$ as well, which is less than 1, of course. it is easier to see the limit using this way, in my humble opinion

**arbolis** · July 11th 2008, 02:04 PM

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

**Jhevon** · July 11th 2008, 02:09 PM

Originally Posted by arbolis

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

yes

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