Dose $\sum_n^{\infty}\frac{z^n}{(\Gamma(z-n+1)n!}$ converge for any values of z other then $0$

Question

We start with

$$\sum_n^{\infty}\frac{z^n}{(\Gamma(z-n+1)n!}$$

we can do a comparison test with $$\sum_n^{\infty}\frac{z^n}{n!}$$ (we could use smaller sires but this makes the algebra easier) witch gives us.

$$|\frac{z^n}{(\Gamma(z-n+1)n!}|<\frac{z^n}{n!}$$ witch gives us. $$|(\Gamma(z-n+1)|>1$$ But know Im Stuck.I have no idea how to evaluate. $\lim_{n \to \infty} (\Gamma(z-n+1)$ And as far as i can tell its indterminit. So what do I do? Is there some algabra thing I could have done?

Answer 1

Throughout this answer, I will consider $\frac{1}{\Gamma(z)}$ as an entire function by resolving all the removable singularities. Then by the Euler's reflection formula, we have

\begin{align*} \frac{1}{\Gamma(1+z-n)} &= \frac{\Gamma(n-z)}{\Gamma(1+z-n)\Gamma(n-z)} \\ &= \frac{\sin\pi(n-z)}{\pi}\Gamma(n-z) \\ &= (-1)^{n-1}\frac{\sin(\pi z)}{\pi}\Gamma(n-z) \end{align*}

which holds a priori for $z \notin \mathbb{Z}$ and then for all $z \in \mathbb{C}$ in the sense of analytic continuation. (Poles of $\Gamma(n-z)$ are cancelled by zeros of $\sin(\pi z)$.) Then by the Stirling's approximation, we know that

$$ \frac{\Gamma(n-z)}{n!} \sim \frac{1}{n^{z+1}} \qquad \text{as } n\to\infty. $$

From this, we find that

• If $z \in \mathbb{Z}$, then all but finitely many terms are zero and the series converges absolutely.
• If $|z| < 1$ then the series converges absolutely.
• If $|z| > 1$ and $z \notin \mathbb{Z}$, all terms are non-zero and the series diverges in view of the ratio test.

(I haven't examined the case $|z| = 1$, though I suspect alternating-series like behavior.) Finally, for $z \in (-1, 0)$ we have

\begin{align*} \sum_{n=0}^{\infty} \frac{z^n}{\Gamma(1+z-n)n!} &= \frac{\sin(\pi z)}{\pi} \sum_{n=0}^{\infty} (-1)^{n-1}\frac{\Gamma(n-z)}{n!}z^n \\ &= \frac{\sin(\pi z)}{\pi} \sum_{n=0}^{\infty} (-1)^{n-1}\frac{z^n}{n!} \int_{0}^{\infty} t^{n-z-1}e^{-t} \, dt \\ &= \frac{\sin(\pi z)}{\pi} \int_{0}^{\infty} \left( \sum_{n=0}^{\infty} (-1)^{n-1} \frac{z^n}{n!} t^{n-z-1}\right)e^{-t} \, dt \\ &= - \frac{\sin(\pi z)}{\pi} \int_{0}^{\infty} t^{-z-1}e^{-(z+1)t} \, dt \\ &= -\frac{\sin(\pi z)}{\pi} \Gamma(-z)(z+1)^z \\ &= \frac{(z+1)^z}{\Gamma(z+1)} \tag{*} \end{align*}

and this extends to all of $|z| < 1$ by analytic continuation. For instance, the following graph is a comparison between the partial sum of first 200 terms (black line) and the function $\text{(*)}$ above (green dashed line):

$\hspace{2em}$

Answer 2

This is not an answer but it (my be) could give you some ideas for your work.

This function is an expansion of a function used in statistical thermodynamics $$f(z)=\sum_{n=0}^\infty \frac{z^n}{\Gamma (z-n+1)\, n!}=\color{red}{\frac{(z+1)^z}{\Gamma (z+1)}}$$ which is defined in the real domain for any $z \geq-1$.

For large values of $z$, using Stirling approximation, it writes $$\log(\left(f(z) \right)=z+1-\log(2\pi z)-\frac{7}{12 z}+\frac{1}{3 z^2}+O\left(\frac{1}{z^3}\right)$$

$f(z)$ goes through a minimum around $z_*\approx -0.6836$ for which $f(z_*)\approx 0.7762$. Inverse symbolic calculators do not recofnize these numbers.