How to calculate this?$S=\sum\limits_{i=1}^\infty(-1)^i(i-1)!x^i$ How to calculate this equation $S=\sum\limits_{i=1}^\infty(-1)^i(i-1)!x^i$ ?
 A: As repeatedly pointed out, your sum $S$ converges (in ordinary sense) only for $x = 0$. This is easily confirmed by the ration test. To be precise, the following theorem will be useful:

Theorem. Assume that $\displaystyle \rho = \lim_{n\to\infty} \frac{|a_{n+1}|}{|a_{n}|}$ exists in $[0, \infty]$. Then the radius of convergence of the power series
  $$ \sum a_n x^n $$
  is equal to $1/\rho$. Here, we adopt the convention that $1/0 = \infty$ and $1/\infty = 0$.

Since there are few clues that restrict the level of background knowledge, there is no reason to stop here. That is, we may consider the sum in some generalized summation sense. In this case, we are going to consider the Borel summation sense. To this end, let
$$\mathcal{B}S(z) = \sum_{n=1}^{\infty} \frac{(-1)^{n} (n-1)!}{n!} z^n = - \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} z^n $$
be the Borel transform of the formal power series $S$. It is clear that this naturally extends to an analytic function for $\Re (z) > -1$ by
$$\mathcal{B}S(z) = -\log (1+z). $$
Thus for $x > 0$, we have
\begin{align*}
S(x)
&= \int_{0}^{\infty} e^{-t} \mathcal{B}S(xt) \, dt
 = \int_{0}^{\infty} e^{-t} \log(1+xt) \, dt\\
&=\left[-e^{-t}\log(1+xt)\right]_{0}^{\infty} + \int_{0}^{\infty} \frac{xe^{-t}}{1+xt} \, dt\\
&=\int_{0}^{\infty} \frac{e^{-t/x}}{1+t} \, dt \qquad (xt \mapsto t) \\
&=e^{1/x} \int_{1}^{\infty} \frac{e^{-t/x}}{t} \, dt \qquad (t+1 \mapsto t) \\
&=-e^{1/x} \int_{-\infty}^{-1/x} \frac{e^{t}}{t} \, dt \qquad (-t/x \mapsto t) \\
&=-e^{1/x} \mathrm{Ei}\left(-\frac{1}{x}\right),
\end{align*}
where $\mathrm{Ei}$ denotes the exponential integral function defined by
$$ \mathrm{Ei}(x) = \mathrm{PV} \!\! \int_{-\infty}^{x} \frac{e^{t}}{t} \, dt. $$
A: This doesn't converge for any $x\neq0$ as the terms do not tend to $0$.
Notice that $n!x^n$ grows arbitrarily large (Take ratios to prove it).
