Convergence of $\sum_{n=1}^{\infty}\frac{n!}{n^n}$

Question

Show whether:

$$\sum_{n=1}^{\infty}\frac{n!}{n^n}$$

converges.

I didn't downvote, but explaining in your question the context (you found that exercise, solved it, and want to write the solution here for the sake of future readers, possibly asking for many different ways to tackle it) would definitely prevent some of the downvotes to come. — Clement C., Dec 06 '17 at 23:29
@ClementC. OK thanks for the feedback. Was just trying to do something positive, seeing that there are several similar (but not equivalent) questions like this on MSE. :/ — Forklift17, Dec 06 '17 at 23:31
That's a worthy endeavor. Just add a notice at the top of your question explaining what you intended to do: without it, your question is virtually indistinguishable from those of the form "I got that homework, do it for me." — Clement C., Dec 06 '17 at 23:33

score 1 · Accepted Answer · answered Dec 06 '17 at 23:17

1

By the ratio test:

$$\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}}\cdot \frac{n^n}{n!} = \left(\frac{n}{n+1}\right)^n \\ =\left(\frac{n}{n+1}\right)^n = \left(\frac{1+n}{n}\right)^{-n} = 1 \div \left(1+\frac{1}{n}\right)^n$$

Taking the limit $n\rightarrow\infty$,

$$1 \div \left(1+\frac{1}{n}\right)^n \rightarrow \frac{1}{e} <1$$

Thus the series converges.

answered Dec 06 '17 at 23:17

Forklift17

317

1

Then why do you ask? – Kenny Lau Dec 06 '17 at 23:19
@KennyLau https://stackoverflow.blog/2011/07/01/its-ok-to-ask-and-answer-your-own-questions/ :) – Forklift17 Dec 06 '17 at 23:26
1

I don't see how this question benefits math.SE as a whole. It isn't like this question is hard, like the situation described in your linked blog. – Kenny Lau Dec 06 '17 at 23:27
1

Standard application of ratio test (AKA bookwork) should not appear as self-questions. – Kenny Lau Dec 06 '17 at 23:27

score 0 · Answer 2 · edited Dec 07 '17 at 02:48

With a big hammer:

By Sterling's approximation: $$ n! \operatorname*{\sim}_{n\to\infty} \sqrt{2\pi n}\left(\frac{n}{e}\right)^n $$ from which $$ \frac{n!}{n^n}\operatorname*{\sim}_{n\to\infty} \sqrt{2\pi} \frac{\sqrt{n}}{e^n} $$ and we can immediately conclude by comparison with the series (with positive terms) $\sum_n \frac{\sqrt{n}}{e^n}$.

(Note that the "simple"-to-prove version of Sterling's suffices, the one with a constant $C>0$ instead of the actual value $\sqrt{2\pi}$.)

Convergence of $\sum_{n=1}^{\infty}\frac{n!}{n^n}$

2 Answers2