[SOLVED] Convergence of an infinite series

**arbolis** · July 10th 2008, 01:19 PM

I've to show whether the following infinite series converges or not. The problem is that I tried to prove it's divergent by induction, by the root test and by trying to bound it from below. But I couldn't reach any conclusion.
Here it comes : $\sum_{n=1}^{+\infty} \frac{n!}{n^n}$ . I'm sure it's a famous one. It's very similar to the one that defines $e^n$ (inversed).
I'd like you to tell me what test should I try. Then if I'm not able to get a conclusion, I'll ask for further help. Thanks in advance.

**Mathstud28** · July 10th 2008, 01:24 PM

Originally Posted by arbolis

I've to show whether the following infinite series converges or not. The problem is that I tried to prove it's divergent by induction, by the root test and by trying to bound it from below. But I couldn't reach any conclusion.
Here it comes : $\sum_{n=1}^{+\infty} \frac{n!}{n^n}$ . I'm sure it's a famous one. It's very similar to the one that defines $e^n$ (inversed).
I'd like you to tell me what test should I try. Then if I'm not able to get a conclusion, I'll ask for further help. Thanks in advance.

Root test...just so you know it is (high-light below)

convergent

Or consider that for very large n we have that (high-light)
n!<<n^n by stirlings approximation

**Plato** · July 10th 2008, 01:25 PM

Does it help to know that $\left( {\sqrt[n]{{\frac{{n!}}{{n^n }}}}} \right) \to \frac{1}{e}$ ?

**arbolis** · July 10th 2008, 01:36 PM

Or consider that for very large n we have that (high-light)
n!<<n^n by stirlings approximation

My intuition fails! I thought that $n!$ would have been much greater than $n^n$ . But from what I did, all seemed to show the contrary and I couldn't believe it.
I was trying to show that if $n^{n+1}+n \geq (n+1)^{n+1}$ , then it would diverges. I recognized it was false, but I wouldn't trust my result!

Does it help to know that

?

Of course! Since $\frac{1}{e}<1$ , the series converges... Now I will try to prove it!

**galactus** · July 10th 2008, 01:39 PM

Plato showed you, but if you wanted you could just use the ratio test:

$\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^{n}}{n!}=\f rac{n^{n}}{(n+1)^{n}}$

${\rho}=\lim\frac{n^{n}}{(n+1)!}=\frac{1}{e}$

${\rho}<1$ ...........convergent

**Mathstud28** · July 10th 2008, 01:49 PM

Originally Posted by galactus

Plato showed you, but if you wanted you could just use the ratio test:

$\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^{n}}{n!}=\f rac{n^{n}}{(n+1)^{n}}$

${\rho}=\lim\frac{n^{n}}{(n+1)!}=\frac{1}{e}$

${\rho}<1$ ...........convergent

Might as well jump on the band-wagon.

$\text{As}\quad{n\to\infty}\quad{n!\sim\sqrt{2\pi{n }}n^ne^{-n}}$

$\lim_{n\to\infty}\sqrt[n]{\frac{n!}{n^n}}\sim\lim_{n\to\infty}\sqrt[n]{\frac{\sqrt{2\pi{n}}n^ne^{-n}}{n^n}}=\frac{1}{e}\lim_{n\to\infty}(2\pi{n})^{\ frac{1}{2n}}$

so using the connection of the root and ratio test

$\lim_{n\to\infty}(2\pi{n})^{\frac{1}{2n}}=\sqrt{\l im_{n\to\infty}(2\pi{n})^{\frac{1}{n}}}=\sqrt{\lim _{n\to\infty}\frac{2\pi(n+1)}{2\pi{n}}}=\sqrt{1}=1$

$\therefore\sqrt[n]{\lim_{n\to\infty}\frac{n!}{n^n}}=\frac{1}{e}$

**galactus** · July 10th 2008, 01:56 PM

We certainly know it converges. Now find what it converges to.

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