Determine whether the sequence is convergent or divergent. If it is divergent, find its limit.

$$ \left\{\frac{n!}{n^n} \right\} $$

I tried to write out some of the terms of this sequence, and this is what I concluded:

$$\frac {(1)(2)(3)\ldots(n)}{(n)(n)(n)\ldots(n)} < 1 $$

I think the sequence converges to zero, but how can I show this?


Hint: If $n$ is even, then half of the fractions $\dfrac{1}{n}, \dfrac{2}{n}, \ldots, \dfrac{n}{n}$ are less than or equal to $\dfrac{1}{2}$ and the other half are less than or equal to $1$. Therefore, $0 \le \dfrac{n!}{n^n} \le \left(\dfrac{1}{2}\right)^{n/2} \cdot 1^{n/2} = \dfrac{1}{2^{n/2}}$.

You can do a similar thing if $n$ is odd.



$$0< \frac {(1)(2)(3)\ldots(n)}{(n)(n)(n)\ldots(n)} < \frac{1}{n}$$



$$a_n=\frac{n!}{n^n}\implies\frac{a_{n+1}}{a_n}=\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e<1\implies \sum_{n=1}^\infty\frac{n!}{n^n}\;\;\text{converges}\implies$$



I'm probably going a little overboard here, but I remember some of the simplifications were confusing to me the first time I learned this. You need to use the ratio test:

$$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}} \right| = L$$

Where: $$ L > 1 \implies \text{Divergent} \\ L < 1 \implies \text{Convergent} \\ L = 1 \implies \text{Inconclusive}$$

Using the ratio test gives:

$$ \left| \frac{n!}{n^{n}} \right| = \lim_{n \to \infty} \left| \frac{(n+1)!}{(n+1)^{(n+1)}} \,\cdot\, \frac{n^{n}}{n!} \right|$$

We know that $n! = 1 \cdot 2 \cdot \ldots \cdot n$. $\,$ Therefore: $\, (n+1)! = 1 \cdot 2 \cdot \ldots \cdot n \cdot (n+1)$. If we factor out $(n+1)$ we are left with $(n+1) \cdot n!$. This allows us to cancel out the $n!$ terms. Using this simplification we can say:

$$ \lim_{n \to \infty} \left| \frac{(n+1)!}{(n+1)^{(n+1)}} \,\cdot\, \frac{n^{n}}{n!} \right| = \lim_{n \to \infty} \left| \frac{(n+1)\cdot n!}{(n+1)^{(n+1)}} \,\cdot\, \frac{n^{n}}{n!} \right| = \lim_{n \to \infty} \left| \frac{(n+1)}{(n+1)^{(n+1)}} \,\cdot\, n^{n} \right|$$

Next, we need to do something with $(n+1)^{(n+1)}$. If we pull out a single $(n + 1)$ term we can reduce the exponent from $(n+1)$ to $n$. This leaves us with $(n+1) \cdot (n+1)^{n}$ , which allows us to cancel out the $(n+1)$ terms. Plugging this in gives us:

$$ \lim_{n \to \infty} \left| \frac{(n+1)}{(n+1)^{(n+1)}} \,\cdot\, n^{n} \right| = \lim_{n \to \infty} \left| \frac{(n+1)}{(n+1)\cdot (n+1)^{n}} \,\cdot\, n^{n}\right| = \lim_{n \to \infty} \left| \frac{n^{n}}{(n+1)^{n}} \right|$$

With equal exponents we can combine $n$ and $n+1$ like so:

$$ \lim_{n \to \infty} \left| \frac{n^{n}}{(n+1)^{n}} \right| = \lim_{n \to \infty} \left| \left( \frac{n}{n+1} \right)^{n} \right|$$

Now we need to manipulate this result to give us something familiar. If we negate our exponent we are left with:

$$\lim_{n \to \infty} \left| \left( \frac{n}{n+1} \right)^{n} \right| = \lim_{n \to \infty} \left| \left( \frac{n+1}{n} \right)^{-n} \right|$$

Expanding the fraction into 2 fractions and simplifying gives us:

$$ \lim_{n \to \infty} \left| \left( \frac{n+1}{n} \right)^{-n} \right| = \lim_{n \to \infty} \left| \left( \frac{n}{n} + \frac{1}{n} \right)^{-n} \right| = \lim_{n \to \infty} \left| \left( 1 + \frac{1}{n} \right)^{-n} \right|$$

Negating our exponent again gives us:

$$\lim_{n \to \infty} \left| \left( 1 + \frac{1}{n} \right)^{-n} \right| = \lim_{n \to \infty} \left| \frac{1}{\left( 1 + \frac{1}{n} \right)^{n}} \right|$$

We know that:

$$ \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n} = e $$

Using this we can say that:

$$ \lim_{n \to \infty} \left| \frac{1}{\left( 1 + \frac{1}{n} \right)^{n}} \right| = \lim_{n \to \infty} \left| \frac{1}{e} \right| < 1 $$

Therefore: $\displaystyle \frac{n!}{n^{n}} $ is convergent

Hope this helps!

  • $\begingroup$ You should note that the ratio test gives convergence if the series $\sum \frac{n!}{n^n}$. Hence, the sequence converges to $0$. $\endgroup$ – PhoemueX Oct 30 '14 at 6:16
  • $\begingroup$ Ahh, good point. I overlooked the fact that it was a sequence and not a series. $\endgroup$ – steveclark Oct 30 '14 at 6:19

A very useful formula is Stirling approximation of $n!$ as Milly commented.

One of the simplest form is given by $$n!\approx n^n \sqrt{2 \pi n} e^{-n}$$ $$\frac{n!}{n^n} \approx \sqrt{2 \pi n} e^{-n}$$

I am sure that you can take from here.

  • $\begingroup$ One can solve the problem this way, but I would argue that this is way to advanced for such a simple exercise. $\endgroup$ – PhoemueX Oct 30 '14 at 6:17

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