Let $n>0$ be a natural number. I am looking for an M(n) of the following function:
$$ f(n) = (1 - n^{-1/4})^n < M(n) $$
I know that if $n$ goes to infinity $f(n)$ goes to $0$. Now, I wonder if there is some simple expression for $M(n)$ that illustrates this. Since there seem to be some relation with the exponential function, I am looking for something like $M(n) = \exp(-g(n))$; for instance, maybe such $g$ may be $\sqrt{n}$, but I am not sure.
We have
$$\log f(n) = n\log (1-n^{-1/4}) = - n \sum_{k=1}^\infty \frac{1}{kn^{k/4}}.$$
In particular, ignoring all but the first term of the series,
$$\log f(n) < -n \frac{1}{n^{1/4}} = - n^{3/4},$$
whence
$$f(n) < \exp \left(-n^{3/4}\right).$$
Using more terms of the Taylor series of the logarithm, we get tighter bounds, e.g.
$$\log f(n) < - n^{3/4} - \frac{1}{2}\sqrt{n} \Rightarrow f(n) < \exp \left(- \left(n^{3/4} + \frac{1}{2}\sqrt{n}\right)\right).$$