For which $x, c$ does $\sum_{n=0}^{\infty}\binom{x}{n}\sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k} |k - c|^{2/3}$ converge?

Question

Consider the series

$$ \sum_{n=0}^{\infty}\binom{x}{n}\sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k}|k - c|^{2/3} $$

where $x, c$ are real numbers and

$$ \binom{x}{n} = \frac{x(x-1)\cdots(x-n+1)}{n!}. $$

Given $c$, for which values of $x$ does the above series converge?

I know that if $x$ is a nonnegative integer, then the outer-series is finite and some combinatorics can provide us with the summation result. The non-trivial cases are those for which $x$ is not an integer.

---

Attempt 1: Try root test and bound $r$ from above.

For the case where $x$ is a non-integer, I would consider using the root test, so I would look at the value of

$$ r = \limsup_{n\rightarrow\infty} \left| \binom{x}{n} \sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k}|k - c|^{2/3} \right|^{1/n}. $$

and determine whether $r < 1$, $r = 1$, or $r > 1$.

My attempt here will be to bound $r$ from above in hopes of finding cases in which we may conclude $r < 1$. Even if this part is successful, this will only give us a partial result to my main question! Also, note that this might not bare any fruit or useful results!

Anyways, I can use the triangle inequality to find

\begin{align*} \left| \binom{x}{n} \sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k}|k - c|^{2/3} \right| &= \left| \binom{x}{n} \right| \left|\sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k}|k - c|^{2/3} \right| \\ &\le \left| \binom{x}{n} \right| \sum_{k=0}^{n} \left|\binom{n}{k}\right||k - c|^{2/3} \\ &\le \left| \binom{x}{n} \right| \cdot (n + |c|)^{2/3}\sum_{k=0}^{n} \binom{n}{k} \\ &\le \left| \binom{x}{n} \right| \cdot (n + |c|)^{2/3}\cdot 2^{n}. \end{align*}

By Euler's reflection formula,

\begin{align*} \binom{x}{n} &= \frac{\Gamma(x+1)}{\Gamma(n+1)\Gamma(x-n+1)} \\ &= \frac{\Gamma(x+1)}{\Gamma(n+1)}\cdot \frac{\Gamma(n-x)\sin[(n-x)\pi]}{\pi} \\ &= \frac{\Gamma(x+1)}{\Gamma(n+1)}\cdot \frac{\Gamma(n-x)\sin[(n-x)\pi]}{\pi} \\ &= \frac{\Gamma(x+1)}{\Gamma(n+1)}\cdot \frac{\Gamma(n-x)\sin(x\pi)}{\pi} (-1)^{n+1} \\ &= \frac{\Gamma(n-x)}{\Gamma(n+1)}\cdot \frac{\Gamma(x+1)\sin(x\pi)}{\pi} (-1)^{n+1} \end{align*}

Next, we invoke the following "Stirling-type bounds" due to Robbins:

$$ \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n} e^{\frac{1}{12n+1}} < \Gamma(n+1) < \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n} e^{\frac{1}{12n}}. $$

With this we have

\begin{align*} \binom{x}{n} &= \frac{\Gamma(n-x)}{\Gamma(n+1)}\cdot \frac{\Gamma(x+1)\sin(x\pi)}{\pi} (-1)^{n+1} \\ &< \frac{\sqrt{2\pi (n-x-1)}\left(\frac{n-x-1}{e}\right)^{n-x-1} e^{\frac{1}{12(n-x-1)}}}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n} e^{\frac{1}{12n+1}}}\cdot \frac{\Gamma(x+1)\sin(x\pi)}{\pi} (-1)^{n+1} \\ &= \left( 1 - \frac{x+1}{n} \right)^{1/2} e^{x+1}e^{\frac{1}{12(n-x-1)} - \frac{1}{12n + 1}}\left(1 - \frac{x+1}{n}\right)^{n}\frac{1}{(n-x-1)^{x+1}}\cdot \frac{\Gamma(x+1)\sin(x\pi)}{\pi} (-1)^{n+1} \end{align*}

Then

\begin{align*} & \left|\binom{x}{n}\sum_{k=0}^{n} \binom{n}{k}(-1)^{n-k}|k - c|^{2/3} \right|^{1/n} \\ &< \left| \left( 1 - \frac{x+1}{n} \right)^{1/2} e^{x+1}e^{\frac{1}{12(n-x-1)} - \frac{1}{12n + 1}}\left(1 - \frac{x+1}{n}\right)^{n}\frac{1}{(n-x-1)^{x+1}}\cdot \frac{\Gamma(x+1)\sin(x\pi)}{\pi} (-1)^{n+1} \right|^{1/n}\left|(n + |c|)^{2/3}\cdot 2^{n}\right|^{1/n} \\ &= \left( 1 - \frac{x+1}{n} \right)^{\frac{1}{2n}+1}e^{\frac{x+1}{n} + \frac{1}{12n(n-x-1)} - \frac{1}{(12n+1)n}}\cdot\frac{1}{(n-x-1)^{\frac{x+1}{n}}} \cdot \left( \frac{|\Gamma(x+1)\sin x\pi|}{\pi} \right)^{\frac{1}{n}}\cdot (n + |c|)^{\frac{2}{3n}}2 \\ &\rightarrow 2. \end{align*}

Thus, by sending $n\rightarrow\infty$, we obtain the upperbound $r\le 2$. Unfortunately, this doesn't tell us any useful info. We want to find out if $r < 1$, $r = 1$, or $r > 1$, and this bound I found doesn't help.

We have to exploit something more subtle about $r$ to prove convergence/divergence. I would like to know any way to progress this as well as any suggestions for alternative avenues. Trying to do numerical experiments in Python with this sum is also tough because it appears to be badly behaved; I cannot even guess when it is convergent or divergent.