Consider the integral $$I(x) = e^{-x} \int_1^x \frac{e^t}t \, dt, \, \, x \to \infty$$
(a) By IBP repeatedly develop a series expansion of the following form, $$I(x)=\bigg( \frac1x +\frac1{x^2} +\frac2{x^3} \bigg) - (1+1+2)e^{1-x}+\dotsb$$
(b) By considering the error in terminating your expansion after $n$ terms, show that the series is asymptotic as $x \to \infty$.
Hint: Show that $|R_n|=O (1/x^{n+1})$ as $x \to \infty$
For (a) I got $$I(x)=\bigg( \frac1x +\frac1{x^2} +\frac2{x^3} \bigg) - (1+1+2)e^{1-x}+\dotsb+ e^{-x} \int_1^x n!t^{-(n+1)}e^t \, dt$$
The last term is $R_n$.
Trying to solve $R_n$ gives $$\bigg[ n!t^{-(n+1)}e^t \bigg]_1 ^x-\int_1^x -(n+1)!t^{-(n+2)}e^t \, dt$$
What do I do now?
What you want to do is produce a bound on the error, rather than evaluate it: the error term is just the difference between the sum and the original integral, after all!
With this idea in mind, we want to show that $$ \lvert R_n \rvert = n! e^{-x} \left\lvert \int_1^x \frac{e^t}{t^{n+1}} \, dt \right\rvert = \int_1^x \frac{e^t}{t^{n+1}} \, dt < \frac{C}{x^{n+1}}. $$ Well, what do we know? We know that $t<x$ everywhere in the integral, by the definition. But that means that $1/x>1/t$, by dividing both sides and using that both are positive. Hence $ 1/x^{n+1} > 1/t^{n+1} $, so $$ n!e^{-x}\int_1^x \frac{e^t}{t^{n+1}} \, dt < n!e^{-x} \frac{1}{x^{n+1}}\int_1^x e^t \, dt, $$ and the remaining integral then evaluates to $e^x-1<e^x$, so the whole lot is $<n!/x^{n+1}$, as we wanted.