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?

  • 2
    $\begingroup$ en.wikipedia.org/wiki/Binomial_series $\endgroup$ May 4, 2017 at 21:06
  • 1
    $\begingroup$ Isn't $2^z$ ?. ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$ $\endgroup$ May 5, 2017 at 6:27
  • $\begingroup$ Felix: It is $2^z$ for the values of $z$ where it converges. Actually, that was the motivation for the question. $\endgroup$ May 5, 2017 at 13:18

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .