which of the following, inequalities is correct?

Question

let n ≥ 3 be an integer . then the statement

$ ({n!}) ^{\frac{1} {n} } $ ≤ $\frac{n+1}{2}$ is

A) true for every n ≥ 3

B)true if and only if n ≥ 5

C) not true for n ≥ 10

D) true for even integers n≥ 6,not true for odd n ≥ 5

my attempts ; i was using the stirling formula $ n! = e^{-n} {n^{n}} $,, now im getting $ ({n!}) ^{\frac{1} {n} } $ $ = (e^{-n} {n^{n}})^\frac{1}{n} $ = $ \frac{n}{e} $ ≤ $\frac{n+1}{2}$..which is true for every n ≥ 3....so correct option is option A).

Is my answer is correct or not .pliz verified and tell me the solution i would be more thankful..

Answer 1

Maybe you're confusing an asymptotic estimate with an equivalence, since the formula you wrote is (obviously) not true: you can simply try it for low values for $n$.

So, the inequality is true for $n=3$, since $6^{\frac{1}{3}}\leq 4$ (you can try to add some details to understand this inequality without any calculation). This fact shows that the B) case is false.

Moreover, if you work by induction, you can see that the option A) is the correct one. In fact:

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

You can provide details for yourself.

Answer 2

By AM-GM $$\frac{n+1}{2}=\frac{1+2+...+n}{n}\geq\sqrt[n]{n!}$$