I am solving for the expansion of the incomplete gamma function $\Gamma (0,z)$. I have arrived at the following:
$$\Gamma (0,z) = \int_z^\infty e^{-x} x^{-1} dx = E_1 (z) = \int_1^\infty e^{-xz}x^{-1} dx$$
where $E_1(z)$ is the exponential integral. I also know that this is equivalent to
$$E_1(z) = -\gamma - \ln z - \sum_{n=1}^\infty \frac{(-1)^n z^n}{n n!}. $$
What I would like to know is if there is any way that the series in this equation can be expanded or its convergence solve such that it is equal to $z + O(z^2)$ where $O$ is the error approximation of the series towards infinity. That is, I need to have
$$E_1(z) = -\gamma - \ln z + z + O(z^2). $$
I also know that the series is a Taylor expansion of some function but I do not yet know what this function is nor if it converges to $z$.
I am quite new to all this and am having a hard time piecing all the information. I would very much appreciate any help. I am using the 1972 Handbook of Mathematical Functions and Arfken, Webber, and Harris's Mathematical Methods for Physicists as references.
The series converges for all complex $z$ by the ratio test. Then write $$ - \sum\limits_{n = 1}^\infty {( - 1)^n \frac{{z^{n} }}{{nn!}}} = z - z^2 \sum\limits_{n = 2}^\infty {( - 1)^n \frac{{z^{n - 2} }}{{nn!}}} . $$ If $\left| z \right| \le K$ with some $K>0$, then $$ \left| {\sum\limits_{n = 2}^\infty {( - 1)^n \frac{{z^{n - 2} }}{{nn!}}} } \right| \le \sum\limits_{n = 2}^\infty {\frac{{K^{n - 2} }}{{nn!}}} \le \sum\limits_{n = 2}^\infty {\frac{{K^{n - 2} }}{{(n-2)!}}} = \sum\limits_{n = 0}^\infty {\frac{{K^n }}{{n!}}} = {\rm e}^K, $$ i.e., $$ - \sum\limits_{n = 1}^\infty {( - 1)^n \frac{{z^{n - 2} }}{{nn!}}} = z + \mathcal{O}(z^2 ) $$ on compact subsets of $\mathbb C$.