Can you use mathematical induction to prove that #5^n < n!# for all #n in ZZ^+, n>=12#?

Question

1337 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-03T07:32:02

See explanation...

Proposition

Let #P(n)# be the proposition:

#5^color(blue)(n) < color(blue)(n)!#

#color(white)()#
Base case

#P(color(blue)(12))# is true since:

#5^color(blue)(12) = 244140625 < 479001600 = color(blue)(12)!#

#color(white)()#
Induction step

Suppose #P(color(blue)(n))# is true:

#5^(color(blue)(n)) < color(blue)(n)!#

Then #P(color(blue)(n+1))# is true:

#5^(color(blue)(n+1)) = 5*5^n < 5*n! < (n+1)*n! = (color(blue)(n+1))!#

#color(white)()#
Conclusion

Having shown:

#{ (P(color(blue)(12))), (P(color(blue)(n)) => P(color(blue)(n+1)) " for " n >= 12) :}#

We can deduce:

#P(n)# for all #n >= 12#