# Inequalities

• Sep 6th 2012, 06:14 AM
GrigOrig99
Inequalities
Can anyone help me confirm if I've got this one correct? Particularly the part at A, where I've defined (k + 2)! = (k + 1)!(k + 2).

Many thanks.

Q.
(n + 1)! > 2n, for n $\displaystyle \in \mathbb{N}$

Attempt: Step 1: For n = 1...
(1 + 1)! = 2! = 2 & 21 = 2
Since 2 > 2, the statement is true for n = 1.

Step2: Assume the statement is true for n = k, i.e. assume (k + 1)! > 2k.
We must now show that the statement is true for n = k + 1,
i.e. ((k + 1) + 1)! > 2k+1 if (k + 2)! > 2k+1
A: (k + 2)! = (k + 1)!(k + 2) > (k + 1)2k...((k + 1)! > 2k...assumed)
If (k + 1)2k > 2k+1, then (k + 2)! > 2k+1
if (k + 1)2k > 2k+1
if (k + 1)2k - 2k+1 > 0
if 2k[(k + 1) - 2] > 0...which is true for k > 1
=> (k + 2)! > 2k+1
Therefore, the statement is true for n = k + 1, if true for n = k. Thus, the statement is true for all n > 2, n $\displaystyle \in \mathbb{N}$
• Sep 6th 2012, 06:44 AM
Plato
Re: Inequalities
Quote:

Originally Posted by GrigOrig99
Can anyone help me confirm if I've got this one correct? Particularly the part at A, where I've defined (k + 2)! = (k + 1)!(k + 2).
Many thanks. Q. (n + 1)! > 2n, for n $\displaystyle \in \mathbb{N}$
Attempt: Step 1: For n = 1...
(1 + 1)! = 2! = 2 & 21 = 2
Since 2 > 2, the statement is true for n = 1.
Step2: Assume the statement is true for n = k, i.e. assume (k + 1)! > 2k.
We must now show that the statement is true for n = k + 1,
i.e. ((k + 1) + 1)! > 2k+1 if (k + 2)! > 2k+1
A: (k + 2)! = (k + 1)!(k + 2) > (k + 1)2k...((k + 1)! > 2k...assumed)
If (k + 1)2k > 2k+1, then (k + 2)! > 2k+1
if (k + 1)2k > 2k+1
if (k + 1)2k - 2k+1 > 0
if 2k[(k + 1) - 2] > 0...which is true for k > 1
=> (k + 2)! > 2k+1
Therefore, the statement is true for n = k + 1, if true for n = k. Thus, the statement is true for all n > 2, n $\displaystyle \in \mathbb{N}$

You tend to make too much of this.
Part of step 2 should note that assuming $\displaystyle K\ge 2$ we have $\displaystyle (K+1)!\ge 2^K$. So you kn ow that $\displaystyle K+2>2$.
Look at $\displaystyle [(K+2)!=(K+2)(K+1)!\ge(2)(2^K)=2^{K+1}$
• Sep 6th 2012, 07:27 AM
GrigOrig99
Re: Inequalities
Ok, so A should be rewritten as: (k + 2)! = (k + 1)!(k + 2) > (k + 2)2k...?
• Sep 6th 2012, 07:33 AM
Plato
Re: Inequalities
Quote:

Originally Posted by GrigOrig99
Ok, so A should be rewritten as: (k + 2)! = (k + 1)!(k + 2) > (k + 2)2k...?

Yes.
• Sep 6th 2012, 07:39 AM
GrigOrig99
Re: Inequalities
Cont. from there:
If (k + 2)2k > 2k+1, then (k + 2)! > 2k+1
(k + 2)2k > 2k+1
if (k + 2)2k - 2k+1 > 0
if 2k[(k + 2) - 2k] > 0...true when k > 1
=> (k + 2)! > 2k+1