Consider the integral $$\int_1^\infty\frac{\exp(-nx)}{x}dx$$
We get:$$\int_1^\infty\frac{\exp(-nx)}{x}dx=n\int_ n^\infty\frac{\exp(-x)}{x}=nE_1(n)$$
My question is, can we approximate this integral with elementary functions for large n?
I've found out that $$\frac {e^{-n}} n> \int_1^\infty\frac{\exp(-nx)}{x}dx$$ As,$$\int_1^\infty\frac{\exp(-nx)}{x}dx<\int_1^\infty{\exp(-nx)}dx$$ Also,$$\int_1^\infty\frac{\exp(-nx)}{x}dx>\sum_{k=1}^{\infty} \dfrac{\exp(-nk)(1-\exp(-n))}{n(k+1)}=\frac{\exp(-n)}{2n}+\mathcal{O}(1)\exp(-2n)$$
Integration by parts does the trick here:
$$\begin{align}n \int_n^{\infty} dx \frac{e^{-x}}{x} &=e^{-n} - n \int_n^{\infty} dx \frac{e^{-x}}{x^2}\\&= e^{-n} -\frac{e^{-n}}{n} + 2 n \int_n^{\infty} dx \frac{e^{-x}}{x^3} \end{align}$$
and so on. The full asymptotic expansion is
$$n \int_n^{\infty} dx \frac{e^{-x}}{x} \sim e^{-n} \sum_{k=0}^{\infty} \frac{(-1)^k \, k!}{n^k} \quad (n \to \infty)$$
Note that the sum is a divergent series for a fixed, finite $n$, which means practically, you evaluate the sum for a given $n$ such that the last term used is a minimum in magnitude.