I'm solving a complexity question where I have:

$$ n!/2^n $$

The goal is to find an upper bound for this.

My idea is using the fact that: $$ n! = O(n^n)$$ $$ n!/2^n = O((n/2)^n) = O(n^n)$$

But is a correct upper bound, and if so is it the tightest upper bound that can be found? I know that n! is asymptotically larger than 2^n, but I'm struggling to do a tighter analysis.

  • 1
    $\begingroup$ Use Stirling’s approximation. $\endgroup$ – Yuval Filmus May 4 '19 at 21:38

Stirling's approximation states that $$ n! \sim \sqrt{2\pi n} (n/e)^n. $$ It follows that $$ n!/2^n \sim \sqrt{2\pi n}(n/2e)^n = \Theta(\sqrt{n}(n/2e)^n). $$

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