Prove that n!>2^n for any n >= 5.

[david90]

prove

n!>2^n for any n >= 5.

I need help.

 

[eltonjohn]

Re: prove

Do you know what mathematical induction is?

if you need to prove that a property p is true .If it is true for p(1) then is also true for p(n+1) assuming that is true for p
So is easier to prove that is true for p(n+1) starting with a supposition of p being true
The idea is that something true can only imply something also true
something false can imply anything (it's like a drunk !)
In other words :
1) p(1) is true .. (in your case p(5) )
2) then if p is true => p(n+1) is also true

so you need to prove that ( n+1)! > 2^(n+1) => (n+1)! > 2. 2^n

I'll leave it to you to do the conclusion !

Cheers

 

[david90]

Re: prove

not much help. I already have what you've wrote before posting.

yes I know induction.

 

[artem]

prove

It is just enough to prove that begining from 5 multipliers values for factor are more than for 2^n expressed as
2^(n-1) * 2 ~ (n - 1)! * n

following must be proved :
2^(n-1) < (n-1)! and 2 < n for n equal or more than 5

 

[klug]

Re: prove

Hi, david90!

It is very easy, I will show it by logical steps

You have n! and 2^n.

n!=1*2*3...*(n-1)*n

2^n=2*2*...*2  n times

substitude your n by (n+1) in your inequality, you will have:

n!*(n+1)>(2^n)*2  (1)

divide it on two inequalities:
n!>2^n                   (2)
(n+1)>2                  (3)

you see, (1) is result of multiplications (2) and (3)

You see, (3) is surely true for every n bigger then 2.

Also, if bigger1>smaller1 and bigger2>smaller2 then (bigger1*bigger2)>(smaller1*smaller2)

So, if n!>2^n is true for some number m bigger then 2, then multiplication of two true inequalities (2) and (3) will be true inequality for this number m. And result will be the proved ineqality (2) but for m+1. So, all numbers bigger then m will keep this (2) true.

 

[waterman]

Re: prove

I think you could change the condition to n >= 4.

Here i my proof to this question (for detail, see the attached PDF file)

In text,

4! = 24 > 16 = 24.
The formula is true for K=4

Suppose the formula is true for n = k
For case n = k + 1
n! = (k + 1)! = (k + 1)*k! > (k + 1)*2^k > 2*2^k = 2^(k+1) = 2^n
The formula is also true for n = k+1
By mathematical induction, the formula is true for every n >= 4.

 

[heo83]

prove

I'm so lazy to download that pdf. After reading klug's proving, I suggest this explaination. It would be simpler though base on klug's idea:

2 >= 2
3 > 2
4 > 2
5 > 2
...
n - 1 > 2 (it's obvious since n >= 5)
n >= 5 > 2.2

Then multiply all the left hand side and all the right hand site we have:
1.2.3...(n-1)n > 2.2.2...2 (there're are n number of 2)
=> n! > 2^n

 

[greek]

prove

Well if it is true for n=5 just make sure it is an increasing sequence

 

[Element_115]

Re: prove

Just set:

n!=2^n

Find the value for "n" ( of course you can't find it "x.x! = DNE").
Then replace the closest value of "n" with "n±1". This will indicate where
to put the ≤,≥ signs. Graph it and it will be obvious. Or take the 1st &2nd
Derivatives to show 1 Extremea, and justify your statements
mathematically.


%%%%%%%%%%%%%%
%MatLab code
clear all;
n=[0:6];
[na,L]=size;
for i=1:L
y1(i)=2^n(i);
if n(i)==0
y2(1)=1;
else
y2(i)=factorial(n(i));
end
end

plot(n,y1,'c*',n,y2);
%%%%%%%%%%%%%%%%
Good Luck