Convergence of $\sum\limits_{n=0}^\infty {\frac{z(z+1)\cdots(z+(n-1))}{n!}} $

Question

The series below converges to a familiar analytic function in some open half plane. Which half plane and which function?

$$\sum\limits_{n=0}^\infty {\frac{z(z+1)\cdots(z+(n-1))}{n!}}=1+z+\frac12z(z+1)+\frac16z(z+1)(z+2)+\dots$$

In class we only cover that the power series converge in a circle of complex plane, so I do not have any clue about the series like this, any helpful hint or advice is welcome! Thanks in advance

Does this answer your question? Convergence of $\sum_{k=0}^\infty {\binom{z}{k}} $ for complex $z$ — Martin R, Dec 14 '19 at 20:51
Well, that problem is z(z+1)(z+2)....., but my question is z(z-1)(z-2)..... — Cooper, Dec 14 '19 at 21:48
@Cooper really? it seems to me, from the current question, that you are asking about $z(z+1)...$ — clathratus, Dec 15 '19 at 00:06
Oh yeah, I am sorry. I mean my question is z+1 but that question is z-1 — Cooper, Dec 15 '19 at 00:12
@Cooper: You are right, this one is slightly different, I have retracted my closing vote. However, the substitution $z = -\alpha$ shows that the series are closely related, and I assume that a similar approach as in https://math.stackexchange.com/a/1576969/42969 should work here as well. — Martin R, Dec 15 '19 at 07:40

score 0 · Answer 1 · answered Dec 15 '19 at 00:34

Hint:

note that, by using the notation of the Rising and Falling Factorials we can rewrite your sum as $$ \eqalign{ & F(z) = \sum\limits_{0\, \le \,n} {{{z\left( {z + 1} \right) \cdots \left( {z + n - 1} \right)} \over {n!}}} = \cr & = \sum\limits_{0\, \le \,n} {{{z^{\,\overline {\,n\,} } } \over {n!}}} = \sum\limits_{0\, \le \,n} {{{\left( {z + n - 1} \right)^{\,\underline {\,n\,} } } \over {n!}}} = \cr & = \sum\limits_{0\, \le \,n} {\left( \matrix{ n + z - 1 \cr n \cr} \right)} = \sum\limits_{0\, \le \,n} {\left( { - 1} \right)^{\,\,n} \left( \matrix{ - z \cr n \cr} \right)} \quad \to \cr & \to \quad \left( {1 - 1} \right)^{\, - \,z} \cr} $$

Convergence of $\sum\limits_{n=0}^\infty {\frac{z(z+1)\cdots(z+(n-1))}{n!}} $

1 Answers1