How to show $ \int_{x}^{\infty} \frac{-24 e^{t}}{t^{5}} d t = o(\left.\frac{6 e^{t}}{t^{4}}\right|_{1} ^{x})$ In the book for Asymptotic Analysis and Perturbation Theory by W. Paulsen, on page 46 it is given that

Find the asymptotic series as $x \rightarrow \infty$ for the integral
$$
\int_{1}^{x} \frac{e^{t}}{t} d t
$$
...
$\begin{aligned} \int_{1}^{x} \frac{e^{t}}{t} d t &=\frac{e^{x}}{x}-e+\left.\frac{e^{t}}{t^{2}}\right|_{1} ^{x}-\int_{x}^{\infty} \frac{-2 e^{t}}{t^{3}} d t \\ &=\frac{e^{x}}{x}-e+\frac{e^{x}}{x^{2}}-e+\left.\frac{2 e^{t}}{t^{3}}\right|_{1} ^{x}-\int_{x}^{\infty} \frac{-6 e^{t}}{t^{4}} d t \\ &=\frac{e^{x}}{x}-e+\frac{e^{x}}{x^{2}}-e+\frac{2 e^{x}}{x^{3}}-2 e+\left.\frac{6 e^{t}}{t^{4}}\right|_{1} ^{x}-\int_{x}^{\infty} \frac{-24 e^{t}}{t^{5}} d t . \end{aligned}$

However, I can't understand how the author argues that
$$
\int_{x}^{\infty} \frac{-24 e^{t}}{t^{5}} d t = o\!\left(\left.\frac{6 e^{t}}{t^{4}}\right|_{1} ^{x}\right),
$$
namely that this series is an asymptotic series.
 A: As I said in the comments, there are serious issues with the derivation in the book. Here is a corrected derivation. Integrating by parts $N$ times gives
\begin{align*}
\int_1^x {\frac{{e^t }}{t}dt} & = \frac{{e^x }}{x} - e + \int_1^x {\frac{{e^t }}{{t^2 }}dt} 
\\ &
 = \frac{{e^x }}{x} - e + \frac{{e^x }}{{x^2 }} - e + 2\int_1^x {\frac{{e^t }}{{t^3 }}dt} 
\\ &
 = \frac{{e^x }}{x} - e + \frac{{e^x }}{{x^2 }} - e + 2\frac{{e^x }}{{x^3 }} - 2e + 6\int_1^x {\frac{{e^t }}{{t^4 }}dt} 
\\ &
 = \frac{{e^x }}{x} - e + \frac{{e^x }}{{x^2 }} - e + 2\frac{{e^x }}{{x^3 }} - 2e + 6\frac{{e^x }}{{x^4 }} - 6e + 24\int_1^x {\frac{{e^t }}{{t^5 }}dt} 
\\ &
 =  \cdots  = e^x \sum\limits_{k = 1}^N {\frac{{(k - 1)!}}{{x^k }}}  + N!\int_1^x {\frac{{e^t }}{{t^{N + 1} }}dt}  + C_N ,
\end{align*}
where $C_N$ is a constant that depends on $N$. Now, by L'Hôpital's rule,
$$
\mathop {\lim }\limits_{x \to  + \infty }  \cfrac{{N!\displaystyle\int_1^x {\cfrac{{e^t }}{{t^{N + 1} }}dt} }}{{\cfrac{{e^x }}{{x^{N + 1} }}}} = \mathop {\lim }\limits_{x \to  + \infty } \cfrac{{N!\cfrac{{e^x }}{{x^{N + 1} }}}}{{\cfrac{{e^x }}{{x^{N + 1} }}}} = N!,
$$
i.e.,
$$
N!\int_1^x {\frac{{e^t }}{{t^{N + 1} }}dt}  = \mathcal{O}\!\left( {\frac{{e^x }}{{x^{N + 1} }}} \right)
$$
as $x\to+\infty$ for any fixed $N\geq 1$. Accordingly,
$$
\int_1^x {\frac{{e^t }}{t}dt}  \sim e^x \sum\limits_{k = 1}^\infty  {\frac{{(k - 1)!}}{{x^k }}} 
$$
as $x\to+\infty$.
