Proof of $\lim_{x\to\infty}\frac{\text{Ei(x)}}{e^x}=0$

Question

I encountered the following limit while doing calculation $$\lim_{x\to\infty}\frac{\text{Ei(x)}}{e^x}=0$$ which is equivalent to $$\lim_{x \to \infty }e^{-x}\sum_{n=1}^{\infty}\frac{x^n}{n·n!}=0$$ and $$\lim_{x\to\infty}\int_0^\infty \frac{e^{-t}}{t-x}dt=0$$ where the integral is undestood as the Cauchy principal value. Because of my lack of experience with the limit calculations of the power series, I lost my way. Any kind of hint or help appreciated.

Answer 1

Let $m$ and $n$ be a positive integers, and restrict $x$ to be positive. First note that $\lim_{x\to\infty}x^m\mathrm e^{-x}=0$, because $x^m\mathrm e^{-x}$ can be expressed as $$\frac1{{\sum_{k=0}^\infty}\dfrac{x^{k-m}}{k!}},$$where every term in the denominator is positive and each of these terms from $k=m+1$ onward grows without bound as $x\to\infty$. A little rearrangement of the series for $\mathrm{Ei}(x)$ gives $$\mathrm e^{-x}\mathrm{Ei}(x)=\sum_{k=1}^{n-1}\frac{x^k\mathrm e^{-x}}{k\cdot k!}+\frac1n\mathrm e^{-x}\sum_{k=n}^\infty\frac{x^k}{(k/n)\cdot k!},$$which, by comparison of the second (infinite) sum with the exponential series, can be seen to be bounded above by $$\sum_{k=1}^{n-1}\frac{x^k\mathrm e^{-x}}{k\cdot k!}+\frac1n.$$By our first-noted result (for $m=1,..., n-1$), we now obtain $$\lim_{x\to\infty}\mathrm e^{-x}\mathrm{Ei}(x)\leqslant\frac1n.$$Since $n$ is arbitrary, the required result follows