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. $n ! \le n^{n}$
$\frac{n^{n}}{n!}=\frac{n}{n} \times\frac{n}{\left(n-1\right)}\times\frac{n}{\left(n-2\right)}\times\ldots\times\frac{n}{1}$
Clearly $ \frac{n^{n}}{n}\ge\,1$
$\therefore n^{n}\ge\,n!$ True
II. $\left(n!\right)^{2} \le\,n^{n}$ is False
III. $10^{n} \le\,n!$
$\frac{n!}{10^{n}}=\frac{n}{10}\times\frac{n-1}{10}\times\frac{n-2}{10}\times\ldots\times\frac{1}{10}$
Given, $n >\,1000$. Clearly, $\frac{n!}{10^{n}}\ge\,1$
Hence, it is also true.
IV. $n^{n} \le\,\left(2n\right)!$
$\frac{2n!}{n^{n}}=\frac{1\cdot2\cdot3\cdot4\ldots n\left(n+1\right)\left(n+2\right)\ldots\left(n+n\right)}{n\times n\times n\ldots n \,times}$
$=\frac{n!\left(n+1\right)\left(n+2\right)\left(n+3\right)\ldots\left(n+1\right)}{n\times n\times n\times\ldots n \, times}$
$=n! \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\left(1+\frac{3}{n}\right)\ldots\left(1+\frac{n}{n}\right)$
Clearly $n \ge\,1$
$\therefore $ It is also true