# Convergence of $\sum\limits_{k=0}^\infty {\binom{z}{k}}$ for complex $z$

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?

• en.wikipedia.org/wiki/Binomial_series May 4, 2017 at 21:06
• Isn't $2^z$ ?. ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$ May 5, 2017 at 6:27
• Felix: It is $2^z$ for the values of $z$ where it converges. Actually, that was the motivation for the question. May 5, 2017 at 13:18

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.