$\sum_{n=1}^\infty \frac{n! e^n}{n^n}$ converges or diverges

Question

i need some advice on how to prove if this series converges or diverges tried using D'Alembert test but i received $1$ so it does not help me. Thanks.

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

$n^n$ grows much faster than $e^n$ because the exponents are the same and $n$ grows while $e$ is constant. — , Sep 03 '18 at 20:26
@Wojowu yes i change it, somehow i forgot the facotorial notation. — Maor Rocky, Sep 04 '18 at 07:44
Using Stirling's Asymptotic Approximation, the terms are asymptotic to $\sqrt{2\pi n}$, so they do not go to $0$. — robjohn, Sep 04 '18 at 08:43
Possible duplicate of Prove that the series $\sum_{n=1}^\infty \frac{e^nn!}{n^n}$ diverges — Arnaud D., Sep 04 '18 at 14:49
@MaorRocky Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE — user, Oct 23 '18 at 20:57

score 3 · Answer 1 · edited Sep 03 '18 at 20:37

The series converges and we can see that in so many ways that is

root test (maybe the simpler)
ratio test (also effective)
limit comparison test (for example with $\sum \frac{1}{n^2}$)
etc.

As an alternative, note that eventually, that is for $n\ge e^2$ we have $$n^n\ge e^{2n}$$ and since eventually $e^n \ge n^3$

$$\frac{ne^n}{n^n}\le\frac{ne^n}{e^{2n}}=\frac n{e^n}\le \frac{n}{n^3}=\frac1{n^2}$$

the given series converges by direct comparison test with $\sum \frac{1}{n^2}$.

zhw. · Answer 2 · 2018-09-03T21:01:25.820

2

Hint: The $n$th term equals

$$e\left(\frac{e}{n}\right )^{n-1}$$

and $e/n \le e/3$ for $n>2.$

edited Sep 03 '18 at 21:01

answered Sep 03 '18 at 20:20

zhw.

105,693

1

slight correction needed: $e/n \le e/3$ for $n>2$ or $e/n < e/3$ for $n>3$ – user Sep 03 '18 at 20:45
Right, thank you. – zhw. Sep 03 '18 at 21:01

score 1 · Answer 3 · answered Sep 03 '18 at 20:21

With some simple manipulations you can show that you can omit the factor $n$ in the numerator (are you able to do this?

Then you are left with $a_n=\frac{e^n}{n^n}$. To see that this sum converges it is enough to see, that these summands are smaller than lets say $\left(\frac{e}{10}\right)^n$, that is they are smaller than the summands of a (converging) geometric sum.

score 0 · Answer 4 · answered Sep 03 '18 at 20:20

0

With ratio test: $$\frac{a_{n+1}}{a_n}=$$ $$\frac{n+1}{n} \frac{e^{n+1}}{e^n} \frac{n^n}{(n+1)^{n+1}}=$$ $$e\frac{n+1}{n}\frac{n^n}{(n+1)^{n+1}}=$$ $$e\frac{n+1}{n}\frac{1}{n+1}\left(\frac{n}{n+1}\right)^n=$$ $$\frac{e}{n}\left(\frac{n}{n+1}\right)^n \to 0$$ Because $\left(\frac{n}{n+1}\right)^n \to \frac{1}{e}$ and $\frac{e}{n} \to 0$

answered Sep 03 '18 at 20:20

Botond

11,938

Hi everyone thanks for the help but the qustion was written wrongly, i added the (!) sign. – Maor Rocky Sep 04 '18 at 07:45

$\sum_{n=1}^\infty \frac{n! e^n}{n^n}$ converges or diverges

4 Answers4