Question

Among the inequalities, 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. $n^n\le \left(2n\right)!$.

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

Answer 1

Answer: (B)

I, III and IV only

(I.) $n^n=n\times n\times \dots n\left(n\text{times}\right)>n\times (n-1)\times (n-2)\cdots \times 1=n!$

$\Rightarrow n^n\ge n!(\because n\ge n-i\forall i\in [0,n]$

(II.) $(n!{)}^2=n^2\times (n-1{)}^2\cdots \times 1^2=(n\times 1)\times (n-1\times 2)\times (n-2\times 3)\dots (2\times (n-1\left)\right)\times (1\times n)$ ... (1)

$n^n=n\times n\times \dots n\left(n\text{times}\right)$ ... (2)

Here, each term except the first and last terms of (1) are greater than those of (2). For example, $2\times (n-1)=2n-2>n\forall n\ge 1000$

$\therefore (n!{)}^2\ge n^n$.

(III.) ${10}^n=10\times 10\times \dots 10\left(n\text{times}\right)$

$n!=n\times (n-1)\times (n-2)\cdots \times 1$

$\because n\ge 1000$, it can be logically argued that in ${10}^n$, only 9 terms are greater in value than terms in $n!$ and that the terms in $n!$ are continuously increasing in value and hence, for $n\ge 1000$, it can be concluded that $n!\le {10}^n$ is false.

(IV.) Considering the proof of part II it can be argued that $\left(2n\right)!$ is greater than $(n!{)}^2$ both have $n!$ but, $\frac{\left(2n\right)!}{n!}=2n\times (2n-1)\cdots \times (n+1)>n!$ as each and every term of LHS is greater than corresponding term of the RHS for $n>1000$.

Hence, it follows that $\left(2n\right)!\ge n^n$