I know generally how to do it i just dont get how to write it in a formal proof. like x^n is always going to be approaching infinity slower than (n!) for x > 1 and for x = 1 its just 1/n! but what about x =< -1? because this doesnt converge to any number. however (n!) will always be greater than |x| soo...
hey wait maybe i could prove that |x^n|/(n!) -> 0 and use the absolute value rule...
well x^n looks like a hell lotta x's multiplied together and n! looks like n(n-1)(n-2)...! and i realise that for any value of x, there exists n s.t n is wayyyyyy greater than x so n! will be wayyy greater than x^n but how do i prove itt... :S
It's impossible to tell you how to prove this without knowing what you have available! In my oppinion, the simplest way is to note that the series, [itex]\sum x^n/n![/itex] converges to [itex]e^x[/itex]. Since it converges, the sequence of terms, [itex]x^n/n![/itex] must converge to 0. But you may not be able to use that.
ive already tried those but we dont have sufficient knowledge to use that stuff yet.
heres what im trying:
Prove that a_n = (x^n/n!) -> 0 for all x
Choose ε > 0. Then there exists for all N ,which belongs to natural numbers, n>N and |a_n| < ε
so, |(x^n/n!)| < ε => |x^n| < n!*ε (since n! is positive the absolute sign is not significant?)
so |x|/[ε^(1/n)] < [n!^(1/n)].
so choose any natural number N s.t. N > |x|/[ε^(1/n)]