2
$\begingroup$

I have to prove that $a_n$ is (strictly) increasing and diverges

$a_n = \sqrt[n]{n!}$ ; n $\in$ $\ \mathbb{N}$

From sequence I see that $a_n$ increasing to infinitive.

$\sqrt[1]{1!}=1 ,\ \sqrt[2]{2!} \approx 1.41, \ \sqrt[3]{3!} \approx 1.81 ,..., \sqrt[n]{n!} $

$\endgroup$
3
$\begingroup$

The sequence $a_n=\sqrt[n]{n}$ is increasing because $a_n\le a_{n+1}$ $$(n!)^{\frac{1}{n}}\le ((n+1)!)^{\frac{1}{n+1}}=((n+1))^{\frac{1}{n+1}}(n!)^{\frac{1}{n+1}}$$ that is $$ n!\le (n+1)^n $$ and this is true because it is simply $$ n(n-1)(n-2)\cdots 3\cdot 2\cdot 1\le \underbrace{(n+1)(n+1)\cdots (n+1)}_{n \text{ times}} $$

Using Stirling's approximation $n!\sim \left(\frac{n}{e}\right)^n$, one has $$ \sqrt[n]{n!}\sim \frac{n}{e}\to \infty $$ so the sequence diverges.

$\endgroup$
3
$\begingroup$

you want to see :

$\sqrt[n]{n!}< \sqrt[n+1]{(n+1)!}$

i.e., $n!<(n+1)!^{\frac{n}{n+1}}$

i.e., $n!^{n+1}< (n+1)!^n$

i.e., $n!*n!*\dots *n! \text{ (n+1 times)} < (n+1)!*(n+1)!,\dots * (n+1)! (\text{ n times})$

i.e., $n! < (n+1)(n+1)\dots (n+1) \text{ (n times)}$

i.e., $n.(n-1).(n-2).\dots 3. 2. 1 \text { (n terms)}< (n+1)(n+1)\dots (n+1) \text{ (n times)}$

you have $ i< (n+1) $ for all $i < n$

Atleat now yous should be able to see that this is true...

$\endgroup$
0
$\begingroup$

$$\left(\frac{n}{2}\right)^{n/2}\leqslant n!\leqslant n^n$$

Sandwich theorem etc.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.