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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.

generic properties
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    Just a single application of L'Hopital's after writing $Ei$ in the integral form. – A.S. Dec 02 '15 at 12:22
  • Thank you. I just came up with that. The singularity in the integral doesn't matter, does it? – generic properties Dec 02 '15 at 12:24
  • \begin{align} e^{-x}\mathrm{Ei}(x)&=\frac{1}{x}\int_{-\infty}^x\frac{e^{t-x}}{t/x},dt\ &=\frac{1}{x}\int_0^\infty\frac{e^{-t}}{1-t/x},dt\ &\sim\frac{1}{x}\int_0^\infty e^{-t}\left(1+\frac{t}{x}+\frac{t^2}{x^2}+\frac{t^3}{x^3}+\dots\right),dt\&=\frac{1}{x}\left(1+\frac1x+\frac2{x^2}+\frac6{x^3}+\dots+\frac{n!}{x^n}+\dots\right)\end{align}

    (in PV sense)

     – r9m Dec 02 '15 at 12:31
  • $$\lim_{x\to +\infty} x^t e^{-x} = 0$$

    No problem with the singularity then.

     – Enrico M. Dec 02 '15 at 12:40
  • @r9m: Your power series (in $1/x$) has radius of convergence $0$. So it does not converge for any real value of $x$. – John Bentin Dec 06 '15 at 11:11
  • @JohnBentin it's the series expansion at $x = \infty$, I should have written $e^{-x}\operatorname{Ei}(x) \sim_{x \to \infty} \frac{1}{x}\left(1+\frac{1}{x}+\frac{2}{x^2}+\cdots \right)$ – r9m Dec 06 '15 at 11:16

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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

John Bentin
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