Difficulties to understand why $a_n= \frac{n!}{n^n}$ does converge

Question

I am having some difficulty grasping this mathematical concept.

Answer 1

If $2 \leq k \leq n$, then $\frac{k}{n} \leq 1$. Hence,

$$\bigg(\frac{2}{n}\bigg)\bigg(\frac{3}{n}\bigg) \cdots \bigg(\frac{n}{n}\bigg) \leq \underset{(n-1)-\mbox{times}}{\underbrace{1\cdot 1\cdots 1}}=1$$ Multiply both sides by $\frac{1}{n}$ and you will get

$$a_n \leq \frac{1}{n},$$ which I guess it is the part that you don't get.

Answer 2

It is the most simple concept.

For N =1, Its 1/1 = 1

For N = 2, its 2*1 / 2*2 = 2/ 4 ~ 0.5

For N = 3, the value is 6/27 ~0.222

for N = 4 The value is 24/256 ~ 0.09375

As you can see, the higher the value of N, the lower the value of the equation n! / n raised to the power of N.

Clearly the value of N is in the opposite direction of the value of the equation. As N Grows, the value decreases. Thats what the theorem says.

Answer 3

If you have three functions $f(x), g(x), \text{ and } h(x)$ such that $g(x)\le f(x)\le h(x)$, then if $$g(c)= h(c)=k (\text{say})$$

Then $f(c)=k$.

You can easily see that because $f(x)$ is always squeezed between $g(x)$ and $h(x)$, i.e., f(x) is always greater than or equal to $g(x)$ and less than or equal to $h(x)$. So if $g(x)=h(x)$ for some value of $x$, $f(x)$ will be equal to that value.

In your particular case, $g(x)=0, f(x)=\frac{x!}{x^x}, h(x)=\frac{1}{x}$. Can you figure out why $\frac{x!}{x^x}\le\frac{1}{x} \ \forall x\in\mathbb {Z^+}$?