Proof by Induction: 0!+1!+2!+...n! < (n+1)!

[cragar]

Homework Statement

Let n be a natural number.

Prove that 0!+1!+2!+...n! < (n+1)!
and my < sign should be less than or equal to.

The Attempt at a Solution

The proof is by induction
it works for 0 because 0! is less than or equal to 1!
now we assume it works for n=k by the induction axiom.
now we see if it works for k+1
0!+1!+2!+...k!+(k+1)!<(k+1+1)!
Now I am not sure if I can do this but I will replace
0!+1!+2!+...k! with (k+1)!
so now I have (k+1)!+(k+1)!<(k+2)!
2(k+1)!<(k+2)!
2(k+1)!<(k+2)(k+1)!
and k+2 will always be bigger or equal to 2 because k is a natural number.
so the inequality is proved by induction.

 

[Dick]

That works fine. Your induction hypothesis is 0!+1!+2!+...k! <= (k+1)!. So 0!+1!+2!+...k!+(k+1)! <= (k+1)!+(k+1)!.

 

[cragar]

ok thanks so everything i did is ok.