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

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

Explanation

Solution

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×(n−1)×(n−2)...................×1n!=n\times \left( n-1 \right)\times \left( n-2 \right)...................\times 1
For example: 5!=5×4×3×2×1=120.5!=5\times 4\times 3\times 2\times 1=120.
Note that we assume 0!=10!=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⇒ac<bda < b;c < d\Rightarrow ac < bd.
Given inequality we need to prove, is written as below:
(n!)2<nnn!<(2n)!{{\left( n! \right)}^{2}}<{{n}^{n}}n!<\left( 2n \right)!
Case-1:- we try to prove (n!)2<nnn!{{\left( n! \right)}^{2}}<{{n}^{n}}n! .
Now let us take the right hand side of inequality, we have: (n(n−1)............1)2{{\left( n\left( n-1 \right)............1 \right)}^{2}} .
Now let us take the right hand side of inequality we get: (n.n.n..........n)(n(n−1)..................1)\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!n! . Right hand side will be equal to the expression nn{{n}^{n}} .
As we wrote n!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,....................,(n−1)<n1 < 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.............(n−1)<nnn.........n.1.2.............\left( n-1 \right) < nnn.........n.
So, (n!)<n2\left( n! \right) < {{n}^{2}} .
So, we proved the inequality (n!)<nn(n!)\left( n! \right) < {{n}^{n}}\left( n! \right) .
Now we need to prove the second inequality given by nn(n!)<(2n)!{{n}^{n}}\left( n! \right) < \left( 2n \right)! .
When we cancel out the n!n! terms this time the terms remaining on right hand side of inequality are 2n,2n−1,2n−2,................,n+1.2n,2n-1,2n-2,................,n+1.
Now, our left hand side of inequality is written as:
n.n........nn.n........n (n times).
Now our right hand side of inequality, is given by:
2n(2n−1)...........(n+1)2n\left( 2n-1 \right)...........\left( n+1 \right) .
We have the inequalities, in terms of n given by, as follows: n<(n+1),n<(n+2),.............n<(2n−1),n<(2n)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 (n.n..........n)<(2n)(2n−1).........(n+1)\left( n.n..........n \right) < \left( 2n \right)\left( 2n-1 \right).........\left( n+1 \right) .
By multiplying n!n! on both sides, we get the inequality nn(n!)<(2n)!{{n}^{n}}\left( n! \right)<\left( 2n \right)!
Here we proved by combining both inequalities proved here, we get (n!)2<nn(n!)<(2n!){{\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 (n!)\left( n! \right) and then multiplying it back into inequality in both cases is very crucial and the best method to solve this inequality.