Question

If $P_n$ denotes the product of first $n$ natural numbers, then for all $n\in N$ .

• A. $P_n\le n^n$
• B. $P_{n+1}\le n^n$
• C. $P_n\le {\left(\frac{n+1}{2}\right)}^n$
• D. None of the above.

Answer 1

The correct option is C $P_n\le {\left(\frac{n+1}{2}\right)}^n$

Given, $n\in N$

Natural numbers are counting numbers whose set is $N=\{1,2,...\}$

Given, $P_n$ is the product of first $n$ natural numbers.

Therefore, $P_n=(1\ast 2\ast ...\ast n)$

We know that, $A.M.\ge G.M.$

$A.M.$ of first $n$ natural numbers =$\frac{1+2+...+n}{n}$

$G.M.$ of first $n$ natural numbers =$\sqrt[n]{n1\ast 2\ast ...\ast n}$

Therefore,

$\frac{1+2+...+n}{n}\ge \sqrt[n]{n1\ast 2\ast ...\ast n}$

$⟹\frac{n(n+1)}{2n}\ge \sqrt[n]{n1\ast 2\ast ...\ast n}$

since, the sum of first $n$ natural numbers is $\frac{n(n+1)}{2}$

$⟹(\frac{(n+1)}{2}{)}^n\ge 1\ast 2\ast ...\ast n$

$⟹P_n\le (\frac{n+1}{2}{)}^n$ (since, $P_n=(1\ast 2\ast ...\ast n)$)