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on our lesson at our university, our professsor told that factorial has these estimates

$n^{\frac{n}{2}} \le n! \le \left(\dfrac{n+1}{2}\right)^{n}$

and during proof he did this

$(n!)^{2}=\underbrace{n\cdot(n-1)\dotsm 2\cdot 1}_{n!} \cdot \underbrace{n\cdot(n-1) \dotsm 2\cdot 1}_{n!}$

and then:

$(1 \cdot n) \cdot (2 \cdot (n-1)) \dotsm ((n-1) \cdot 2) \cdot (n \cdot 1)$

and it is equal to this

$(n+1)(n+1) \dotsm (n+1)$

why it is equal, I didn't catch it. Do you have any idea? :)

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  • $\begingroup$ sorry it has to be $\left(\frac{n+1}{2}\right)^{n}$ $\endgroup$ – MatFyzak Dec 12 '15 at 13:15
  • $\begingroup$ it must be inequality, not equality. Use geometric vs arithmetic mean. Certainly $(1\cdot2)\dot(2\cdot1)$ is not the same as $3\cdot3$, don't expect that. Perhaps you did not pay attention, perhaps your professor was not clear, either way you should be able to recover and figure out what was meant. $\endgroup$ – Mirko Dec 12 '15 at 13:24
  • $\begingroup$ You can have a look here or here. And some proofs of the other inequality can be found in this question and other posts linked there. $\endgroup$ – Martin Sleziak Oct 21 '16 at 15:40
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Not equal, but by a standard inequality: $\sqrt{ab} \le \frac{a+b}{2}$, so $ab \le \frac{(a+b)^2}{4}$.

So all products $i\cdot ((n+1)-i)$ are estimated above by $\frac{(i + ((n+1)-i))^2}{4} = \frac{(n+1)^2}{4}$. So no equality, but upper bounded by. Because we have $(n!)^2$ we get rid of the square roots again so $n! \le (\frac{n+1}{2})^n$ (where the $n$-th power comes form the fact that we have $n$ terms in the $(n!)^2$ expression.

The first equality is just rearranging terms.

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$$n^{n/2}\le n!\le \left(\frac{n+1}{2}\right)^{n}$$

$$\iff n^n\le (n!)^2\le \left(\frac{(n+1)^2}{4}\right)^n$$

Now, $n\le i((n+1)-i)$ for all $i\in\{1,2,\ldots,n\}$, because this is equivalent to $n(1-i)\le i(1-i)$. If $i=1$, then it's true. If $i\neq 1$, then this is equivalent to $n\ge i$, which is true.

Also $ab\le \frac{(a+b)^2}{4}$ for all $a,b\in\mathbb R$, because this is equivalent to $(a-b)^2\ge 0$, which is true.

$$n^n=\prod_{i=1}^n n\le \underbrace{\prod_{i=1}^n i((n+1)-i)}_{(n!)^2}\le \prod_{i=1}^n\frac{(i+((n+1)-i))^2}{4}=\left(\frac{(n+1)^2}{4}\right)^n$$

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