# Sequence (n!)/(n^n) Convergent or Divergent and Limit?

## Homework Statement

Is the sequence {(n!)/(n^n)} convergent or divergent. If it is convergent, find its limit.

## Homework Equations

Usually with sequences, you just take the limit and if the limit isn't infinity, it converges... That doesn't really work here. I know I'm supposed to write out the terms, but I can't figure out where to go from there.

## The Attempt at a Solution

(n!)/(n^n) = ((1)(2)(3)...(n-1)(n))/((n)(n)(n)...(n)(n))
I think I'm supposed to say that this is less than something that goes to zero, but I don't know what that something should be.

n/n = 1

Does this mean that I can cancel that out and have a list of numbers smaller than n divided by n? Then the sequence is less than 1...but I think it converges to zero, I just don't know how to show that.

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RUber
Homework Helper
You are on the right track. Try writing out a few terms to see if you can see the growth factor from $\frac{n!}{n^n}$ to $\frac{(n+1)!}{(n+1)^{(n+1)}}$.

Okay, so...

((n+1)!)/((n+1)^(n+1)) = (1/(n+1)) ((n+1)!)/((n+1)^n) = (1/(n+1)) (((1)(2)(3)...(n)(n+1))/((n+1)(n+1)(n+1)...(n+1)(n+1)) < (1/(n+1)) and the limit of 1/(n+1) as n goes to infinity is zero...

Is my inequality true? My logic was just that the huge set of terms being multiplied by each other is still less than 1, so a (1/(n+1)) is going to be greater than itself multiplied by a number less than one. Is this correct?

How exactly does this relate to n!/(n^n)?

haruspex
Homework Helper
Gold Member
RUber is suggesting you consider the ratio between consecutive terms, $\frac {n!}{n^n}$ and $\frac {(n+1)!}{(n+1)^{n+1}}$.

Zondrina
Homework Helper
The ratio test would work I believe. In general though: $n^n > n! > b^n > n^x > \ln(n)$ as $n \rightarrow \infty$. So its easy to determine convergence by merely examining the behaviour of $a_n$.

RUber
Homework Helper
I think you have said it for the n+1 case. The logic holds for the nth case just the same.
$\frac{n!}{n^n}=\frac1n \frac2n ... \frac nn$ the largest term in the product is 1, so the product can be no larger than the smallest term. Just as you said in post #3.