1/(2^n)! is negligible function?

Question

By definition $\frac {1} {n}$, $\frac{1}{2^n}$ and $\frac{1}{n!}$ are negligible functions.

I have got the function $$f(n) = \frac{1}{(2^n)!}$$ where $n$ is security parameter.

I don't understand, How do I formally proof that $$\frac{1}{(2^n)!}$$ is a negligible function? Can anyone please help me out?

You may want to note that $1/n$ is not a negligible function because $1/n^2$ approaches $0$ much faster... — SEJPM, Aug 23 '19 at 17:52
Possible duplicate of Negligible function defined as product of polynomial and a negligible function — kelalaka, Aug 23 '19 at 23:07

score 1 · Accepted Answer · answered Aug 23 '19 at 18:00

The simplest way to prove a function $f$ is negligible if it is obviously negligible is to show that it is "more negligible" than some other function $g$ which you have already proven negligble, e.g. $g(n)=2^{-n}$.

Because $g$ is negligble, there exists $n_{g_0}$ such that for all $n>n_{g_0}$ it holds that $g(n)<1/{n^c}$ for any fixed choice of $c$.

Now you can capitalize on that by showing that there exists some $n_{f_0}$ such that for all $n>n_{f_0}$ it holds that $f(n)\leq g(n)$. Clearly it then holds that for all $n>\max(n_{f_0},n_{g_0})$ that $f(n)\leq g(n)<1/n^c$ which means $f$ is negligible.

I'll leave the choice of $g$ for your concrete application to you.

I, usually, find the limit definition much easier to play. – kelalaka Aug 24 '19 at 20:40 — kelalaka, Aug 24 '19 at 20:40

1/(2^n)! is negligible function?

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