Question

Is n!/(n^n) a decreasing function.

Answer 1

if you expand it,
(n* n-1 * n-2 ... ) / ( n*n*n*n .. )

Answer 2

How do you show that it is decreasing

Answer 3

I believe that how you do this in general is you show for two random numbers P and Z, where P>Z, and your function F(x), F(P) < F(Z). Am I correct?

Answer 4

Yep.
You can prove it by noting that if \(0<A<B\), then:
\[
0<\frac{A}{B}<1
\]And, if:
\[
\frac{A}{B}<1\implies\\
\frac{A}{B}\cdot\frac{C}{D}<\frac{C}{D}
\]For \(C,D > 0\).
It's actually a pretty simple, nice proof.

Answer 5

I think it can not be determine cause it can not be used by 'Hospital's Theorem。

Answer 6

^ No necessity, there's an elementary proof.

Answer 7

really? but how to prove?

Answer 8

check the ratio of one term to the next

Answer 9

\(a_n>a_{n+1}\) check if \(\frac{a_{n+1}}{a_n}<1\)

Answer 10

All right, so we have that, we need to prove that, for \(N>1\) (where \(N\in\mathbb{Z}\)):
\[
\frac{(N+1)!}{(N+1)^{N+1}}-\frac{N!}{N^N}<0
\]It suffices to prove that:
\[
\frac{(N)!}{(N)^{N}}<\frac{(N-1)!}{(N-1)^{N-1}}
\]All right, recall the definition of factorial, and of integer exponentiation:
\[
\frac{N(N-1)(N-2)\cdots(2)(1)}{\underbrace{N\cdot N \cdot N \cdots N}_{N}}
\]Note that this becomes:
\[
\frac{N}{N}\frac{N-1}{N}\frac{N-2}{N}\frac{N-3}{N}\cdots\frac{1}{N}
\]But, since:
\[
N-1<N
\]Then:
\[
\frac{N-1}{N}<1
\]So:
\[
P\frac{N-1}{N}<P
\]For any \(P>1\).
Assume that \(P=\frac{(N-2)!}{(N-2)^{N-2}}\)
Then:
\[
\frac{N-1}{N}\frac{(N-2)!}{(N-2)^{N-2}}<\frac{(N-2)!}{(N-2)^{N-2}}
\]Which means:
\[
\frac{(N-1)!}{(N-1)^{N-1}}<\frac{(N-2)!}{(N-2)^{N-2}}
\]Which is what we wished to prove.

Answer 11

Wait, missing some step in my proof, there's a comparison on a per-term case, that I'm missing! Bah, sorry.

Answer 12

Aka, also note that \(N^N>(N-2)^{N-2}\), hence:
\[
\frac{1}{N^N}<\frac{1}{(N-2)^{N-2}}
\]