Taylor series without factorial term We all know and love the fact that in most of cases  $ f(z)=\sum_{n=0}^{\infty}\frac { f^{(n)}(0)z^n}{n!} $
In context of my question, what I need is general method of evaluation  $\sum_{n=0}^{\infty}f^{(n)}(0)z^n$
as transformation of $f (z) $
 A: First let's find transformation that $g (t^n) \rightarrow n!z^n $
$\displaystyle \int_0^{\infty} \frac {t^n e^{-\frac {t}{z}}}{z}dt =n!z^{n}$
By now, solution is trivial.
$\displaystyle \sum_{n=0}^{\infty}f^{(n )}(0)z^n= \sum_{n=0}^{\infty}\frac{f^{(n)}(0)z^n n! }{n!}=\sum_{n=0}^{\infty}\frac{f^{(n)} (0)  \int_0^{\infty} \frac {t^n e^{-\frac {t}{z}}}{z}dt }{n!}=\int_0^{\infty} \frac {  f (t) e^{-\frac {t}{z}}}{z} dt $
Hypothesis: If we just assume that integral at infinity is convergent to zero (Convergence of definite integral and its analytic continuation), we get
$\displaystyle \sum_{n=0}^{\infty}f^{(n)} (0)z^n=-\lim\limits_{t \rightarrow 0}\int \frac {  f (t) e^{-\frac {t}{z}}}{z} dt $
Edit:I've just realise, that I have discovered Laplase transform XD
A: I've got $3$ examples for you in $\mathbb C$, but they only apply for $|z| < 1$:
$$\sum_{n=0}^{+\infty} z^n = \frac{1}{1-z} \tag 1$$
$$\sum_{n=0}^{+\infty} (-1)^n z^n = \frac{1}{1+z} \tag 2$$
$$\sum_{n=0}^{+\infty}(-1)^n z^{2n} = \frac{1}{1+z^2} \tag 3$$
Note that they also work in $\mathbb R$, but you didn't specify the nature of your "$x$".
