Proving Lim (2^n)/sqrt(n!) = 0

[transfear]

So I've been asked to prove that:

lim (n-->infinity) [2^n]/sqrt(n!) = 0

I've tried fiddling with Stirling and L'Hospital, but can't find my way through it.

Any thoughts?

 

[Dick]

Show us some of your 'fiddling'. If you use Stirling, it's not that hard.

 

[transfear]

Stirling brought me to:

lim n-->inf [2^n * e^(n/2)]/n^(n/2+1/4)

I've also been able to express it as:

lim n-->inf [2 * e^(1/2)]^n / n^(n/2+1/4)

And I won't show you L'Hospital's results since it gets horrible... Anyways, I feel there's something obvious about it that I cannot see. L'Hospital won't help me as long as I have something^n in my numerator...

I'm turning in circle.

 

[Dick]

Look at the log of your expression. log(2^n/sqrt(n!))=log(2^n)-log(sqrt(n!))~n*log(2)-(1/2)*n*log(n). That goes to -infinity as n->infinity, right? That means 2^n/sqrt(n!) goes to zero, also right? Notice I threw out all terms in Stirling's approximation except for log(n!)~n*log(n). The wouldn't be important unless the n*log(n) had canceled out. It grows faster than all the other terms.

 

[transfear]

Thanks for the hint, but I think I may have found another way to prove it.

2^n may be splitted into n-1 products of n: 2 * 2 * 2 * 2 * 2 * ... * 2 <-- doing this n times
sqrt(n!) may also be splitted into products like: sqrt(1) * sqrt(2) * sqrt(3) * ... * sqrt(n)

Thus, I may recreate products of fraction like:
[2/sqrt(1)] * [2/sqrt(2)] * [2/sqrt(3)] * [2/sqrt(4)] * ... [2/sqrt(n)]

This makes it obvious that the limit converges to 0 since denominators increase, but numerators stay constant. Right?

 

[Dick]

Sure. That's another way. To phrase it differently, if f(n)=n!/sqrt(n!) then f(n+1)/f(n)=2/sqrt(n+1). That ratio is less than 1/2 for large n. You can appeal to sort of a ratio test.