Question

Among the inequalities below, which ones are true for all natural numbers $n$ greater than $1000$ ?
I. $n!\le n^n$
II. $(n!{)}^2\le n^n$
III. $10^n\le n!$
IV. $n^n\le (2n)!$

• A. I and IV
• B. I, III and IV
• C. II and IV only
• D. I, II, III and IV

Answer 1

Answer: (B)

We have,
I. $n!\le n^n$
$\frac{n^n}{n!}=\frac{n}{n}\times \frac{n}{\left(n-1\right)}\times \frac{n}{\left(n-2\right)}\times \dots \times \frac{n}{1}$
Clearly $\frac{n^n}{n}\ge 1$
$\therefore n^n\ge n!$ True
II. ${\left(n!\right)}^2\le n^n$ is False
III. $10^n\le n!$
$\frac{n!}{10^n}=\frac{n}{10}\times \frac{n-1}{10}\times \frac{n-2}{10}\times \dots \times \frac{1}{10}$
Given, $n>1000$. Clearly, $\frac{n!}{10^n}\ge 1$
Hence, it is also true.
IV. $n^n\le \left(2n\right)!$
$\frac{2n!}{n^n}=\frac{1\cdot 2\cdot 3\cdot 4\dots n\left(n+1\right)\left(n+2\right)\dots \left(n+n\right)}{n\times n\times n\dots ntimes}$
$=\frac{n!\left(n+1\right)\left(n+2\right)\left(n+3\right)\dots \left(n+1\right)}{n\times n\times n\times \dots ntimes}$
$=n!\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\left(1+\frac{3}{n}\right)\dots \left(1+\frac{n}{n}\right)$
Clearly $n\ge 1$
$\therefore$ It is also true