Question

Discretemath: Prove 2^n < (n+1)! for n>=2

Answer 1

this is equivalent to the statement
2^(n+1)<(n+2)! for n>=1. Now we prove by induction

let n=1,
2^2=4<6=3!
so it is true when n=1.
Now suppose it is true for a natural number n>=1.
then 2^(n+1)<(n+2)!

suppose for contradiction,
2^(n+2)>=(n+3)!
=> 2^(n+1)>=(n+3)!/2=(n+3)(n+2)!/2>(n+3)2^(n+1)/2=(n+3)2^2^n
=> 2^(n+1)>=(2^n)(n+3)
since n>=1,
=>2^(n+1)>=(2^n)(n+3)>=(2^n)4=(2^n)2^2
=> 2^(n+1)>=2^(n+2) <-- a contradiction since n>=1.

hence 2^(n+1)<(n+2)! for n>=1 which implies 2^n < (n+1)! for n>=2. Ther is no way to prove this for real numbers since (n+1)! would be undefined if n+1 would not be a natural number

Answer 2

lol that s not how u do proof by induction and u cant just change the statement u r proving

Answer 3

This is wat i did

Proof by induction
Base step: let P(2)
2^2<(2+1)!=4<3*2*1=4<6
Therefore, the claim holds for n=2

Induction step: let P(n+1)
2^n+1 < (n+1+1)!
2^n*2 < (n+2)!
2^n *2 < n! * n+1 * n+2

Answer 4

and obviously n+1 and n+2 is > 2 since n>=2
now prove 2^n < n!

Answer 5

\[2^n=2\times2\times2\times\cdots \times 2<2\times 3\times 4\times\cdots(n+1)=(n+1)!\]

Answer 6

Maybe like this is more clear:
(1)For n=2, the statement is valid.
(2)For n>=2 the statement is 2^n<(n+1)!
(3)For n+1, the statement gets: 2^n*2<(n+2)(n+1)!
Since step 2 implies that (n+1)!/2^n>1, then by (3) 2/(n+2)<(n+1)!/2^n.
For the statement to be true, 2 must make 3 true, and 2/(2+n) < 1. Since this last statement is true, by induction, the original statement is true.

Answer 7

@Zarkon i know this is the logic, but is that considered a proof?

Answer 8

I might add a few words...about how the for each 2 there is a number in the factorial that can be matched to it that is bigger than or equal to two...therefore the RHS must be bigger

Answer 9

I think induction is a little overkill for this problem