Divergency of the asymptotic expansion of $\operatorname{Ei}(x)$

Consider function $f(z)=e^{-z}\int^z\frac{e^t}tdt$, it satisfies the differential equation $w'+w=\frac1z$. Let $(w')$'s asymptotic expansion as $z\to\infty$ be $w'\sim Ce^{-z}+\sum\limits_{n=1}^\infty\frac{a_n}{z^n}$. Substitute it into $w'+w=\frac1z$ gives $w\sim Ce^{-z}+\sum\limits_{n=1}^\infty\frac{(n-1)!}{z^n}$.
My questions
(i)If I let $w'=Ce^{-z}+\sum\limits_{n=1}^\infty\frac{a_n}{z^n}$, I won't get a result because of the divergency of the series. Why does the series diverge?
(ii)Is there a way to determine whether the asymptotic expansion of $\int f(z)dz$ or a differential equation's solution converge or diverge?
Let $t=\frac1z$, $t\to0$. $\frac{dw}{dt}-\frac1{t^2}w=-\frac1t$. $P(t)=-\frac1{t^2}$ and $Q(t)=-\frac1t$. They have poles at $t=0$. Probably that is why the asymptotic expansion diverges?

• It diverges because $\frac{(n-1)!}{z^n}$ does not go to zero for any $z$ if $n\rightarrow \infty$. The summands are essentially the inverse of the terms of the Taylor series for $e^x$. – gammatester Sep 4 '18 at 9:14
• @gammatester Yes, I am trying to find an explanation from the integral and differential equation themselves. – Kemono Chen Sep 4 '18 at 9:22