Determine if the series a(n) = n! / 2^n converges or diverges

I know the answer to this problem, the answer is the series diverges because the bottom factor increases at a rate faster then the top factor for large values of n.

My question is, HOW DO I PROVE THAT THE NUMERATOR INCREASES AT A FASTER RATE THAN THE DENOMINATOR AT LARGE VALUES OF N.

Sorry to put my question in all caps, but the majority of the time I post stuff on here, people seem to always misread my questions and give me off the wall answers.

Stirling's formula: n! ~ (n/e)^n sqrt(2pi n)

This gives you a good estimate of n! & indicates the rate of increase.

Look this up in Wikipedia. Or google "Stirling's formula".
Or "gamma function", Euler's generalization of n! to all real numbers.

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Simpler but much less accurate: N(n) = n!, N(n+1) = (n+1) N(n); N(n+1) - N(n) = n N(n)

This is like the differential equation dN(t)/dt = t N(t)

D(n) = 2^n = e^(an), where a = ln(2)

Obviously define D(t) = 2^t = e((at)

d/dt D(t) = a D(t)

In ist case rate of increase = tN(t); 2nd case rate of increase = a D(t)

t increases while 2 is constant.

This is a textbook example of a situation where you'd use the ratio test:
Start by taking the limit as n->∞ of a(n+1)/a(n)
a(n+1) = (n+1)!/2^(n+1)
a(n) = n!/2^n

divide the two and you get:

ratio(n) = (n+1)/2

taking the limit as n --> infinity and you have

ratio = infinity

since the (limit of the) ratio is greater than 1, you can safely state that the series diverges.

P.S. there's a reason they don't give you a derivative of the factorial function. First off, it isn't implicitly defined to be a continuous function. Second, if you went through the trouble to find what a continuous version of the factorial would look like (see the "Gamma Function"), you find out that the derivative of the "factorial" function would be

y' = integral (from 0 to infinity) of [t^x * e^-t * lnt]dt

because sometimes math is a cruel mistress

i think numerator will be much greater number than denominator (as n approaches to infinity)
bottom is only 2's ( 2*2*2... 100 times, for instance)
top is, for instance, 100*99*98*97 ...

1st : use the Ratio Test for the divergence

2nd : f(n) = n! / 2^n < f(n+1) = [ n+1]! / 2^(n+1) for all n > 0

and your 1st statement is in error { see your ' question' }