For what complex values of $z$ does the following sum converge?
$$\sum_{k=0}^\infty {\frac{z(z-1)\cdots(z-k+1)}{k!}} $$
And how would you prove it?
Mathematica seems to suggest the sum converges as long as $\Re(z) \ge 0$, regardless of the imaginary part. Is that right?
According to the Wikepedia article Binomial series, which is about the series $$ (1 + x)^\alpha = \sum_{k=0}^\infty {\alpha \choose k} x^k \tag{1}$$ if $\,x=1\,$ then $\,|x|=1\,$ and the convergence is
If $\Re(\alpha)>0$, the series converges absolutely.
If $-1<\Re(\alpha)\le 0$, the series converges conditionally if $x\ne -1$ and diverges if $x=-1.$
If $\Re(\alpha)\le -1$, the series diverges.
In the case in question $$ \sum_{k=0}^\infty {z \choose k} \tag{2}$$ this translates to if $\Re(z)>0$ the convergence is absolute, if $-1<\Re(z)\le 0$ the convergence is conditional., and if $\Re(z)\le -1$ the series diverges.
The summary is that if $\Re(z)>-1$ the series converges, otherwise it doesn't. Notice, that unlike power series, binomial series such as $(2)$ do not have a radius of convergence.
