Question

Prove that ${{\left( n! \right)}^{2}}<{{n}^{n}}n!<\left( 2n \right)!$ for all positive integers n.

Answer 1

Hint: First look at the definition of factorial. Break this one inequality of 3 terms into 2 inequalities each containing 2 terms. Now, expand the both sides of each inequality and compare the terms to prove the inequality. Use the known general condition said as follows: “If two or more terms are multiplied we get a result which is greater than all terms multiplied.”

Complete step by step answer:
Factorial:- In mathematics, factorial is an operation, denoted by “$\left( ! \right)$ “. It represents the product of all numbers between 1 and a given number.
In simple words factorial of a number can be found by multiplying all terms you get by subtracting 1 from a given number repeatedly till you get the number difference as 1. Its representation can be written as: $n!=n\times \left( n-1 \right)\times \left( n-2 \right)...................\times 1$
For example: $5!=5\times 4\times 3\times 2\times 1=120.$
Note that we assume $0!=1$ . It is standard value. It has a wide range of applications in combinatorics.
Given a condition on n is that n is always positive.
So, we can say that if $a < b;c < d\Rightarrow ac < bd$.
Given inequality we need to prove, is written as below:
${{\left( n! \right)}^{2}}<{{n}^{n}}n!<\left( 2n \right)!$
Case-1:- we try to prove ${{\left( n! \right)}^{2}}<{{n}^{n}}n!$ .
Now let us take the right hand side of inequality, we have: ${{\left( n\left( n-1 \right)............1 \right)}^{2}}$ .
Now let us take the right hand side of inequality we get: $\left( n.n.n..........n \right)\left( n\left( n-1 \right)..................1 \right)$
So, by cancelling common terms we get that left hand side will be equal to the expression $n!$ . Right hand side will be equal to the expression ${{n}^{n}}$ .
As we wrote $n!$ , we can say that n is greater than all numbers between 1 to n.
So, we write the inequalities given by (in terms of n):
$ 1 < n,2 < n,....................,\left( n-1 \right) < n$ .
By multiplying all inequalities, given n is positive so we can apply this, we get: $1.2.............\left( n-1 \right) < nnn.........n.$
So, $\left( n! \right) < {{n}^{2}}$ .
So, we proved the inequality $\left( n! \right) < {{n}^{n}}\left( n! \right)$ .
Now we need to prove the second inequality given by ${{n}^{n}}\left( n! \right) < \left( 2n \right)!$ .
When we cancel out the $n!$ terms this time the terms remaining on right hand side of inequality are $2n,2n-1,2n-2,................,n+1.$
Now, our left hand side of inequality is written as:
$n.n........n$ (n times).
Now our right hand side of inequality, is given by:
$2n\left( 2n-1 \right)...........\left( n+1 \right)$ .
We have the inequalities, in terms of n given by, as follows: $n<\left( n+1 \right),n < \left( n+2 \right),.............n < \left( 2n-1 \right),n<\left( 2n \right)$ .
As n is given as positive, we can multiply these inequalities $\left( n.n..........n \right) < \left( 2n \right)\left( 2n-1 \right).........\left( n+1 \right)$ .
By multiplying $n!$ on both sides, we get the inequality ${{n}^{n}}\left( n! \right)<\left( 2n \right)!$
Here we proved by combining both inequalities proved here, we get ${{\left( n! \right)}^{2}} < {{n}^{n}}\left( n! \right) < \left( 2n! \right)$ .


Note:
The idea of multiplying inequalities will be applicable only when n is positive. Students confuse and multiply at all times and get wrong results. The idea of cancelling $\left( n! \right)$ and then multiplying it back into inequality in both cases is very crucial and the best method to solve this inequality.