Strange limit problem involving $\binom{z}{n} e^{-xn\log n}$ with $z \in \mathbb{C}$ Is the following limit result correct: $$\lim\limits_{x \to 0^{+}}\sum\limits_{n=0}^{\infty} \binom{z}{n} e^{-xn\log n} = 2^{z}$$ where, $z \in \mathbb{C}$, and the notation $\displaystyle \binom{z}{n} = \frac{1}{n!}\prod\limits_{k=0}^{n-1}(z-k)$
I verified the result for $z = -1$ case and the limit is indeed $\dfrac{1}{2}$ in that case. Not sure how to proceed with the general complex $z$ case. Any suggestions are welcome. Thanks.
(It would be great if someone could verify the result with a mathematical software)
 A: If $\Re z>0$, the series $\sum_n \binom{z}n$ converges absolutely (by Raabe test, for example), thus we may replace $x$ to 0 and just need to check that this sum equals $2^z$. This follows from another relaxation: $\sum \binom{z}{n}t^n=(1+t)^z$ for $0<t<1$ and we may let $t$ tend to 1.
Next, consider the case $-1<\Re z\leqslant 0$. For $\alpha_n:=e^{-xn\log n}$ denote $\beta_n=\alpha_n-\alpha_{n+1}+\dots$, then $\alpha_n=\beta_{n}+\beta_{n+1}$ and we have $$\sum \binom{z}{n}\alpha_n=\sum \binom{z}{n}(\beta_n+\beta_{n+1})=\sum \binom{z+1}n\beta_n.$$
Now $\beta_n$ are uniformly bounded, since $\alpha_n$ are decreasing, and the series $\binom{z+1}n$ already absolutely converges (to $2^{z+1}$). Therefore it suffices to prove that for each fixed $n$ we have $\beta_n\rightarrow 1/2$ when $x\rightarrow +0$. We have
$$
\beta_n=\frac12 \alpha_n+\frac12 \sum_{k\geqslant n,k\equiv n\pmod 2} (\alpha_k-2\alpha_{k+1}+\alpha_{k+2}).
$$
The first summand tends to $1/2$, of course, and I claim that the sum tends to 0. Each summand does tend to 0, so it suffices to majorate them by a convergent series not depending on $x$. We have
$$
\alpha_k-2\alpha_{k+1}+\alpha_{k+2}=w''(k+\theta),
$$
where $w(t)=e^{-xt\log t}$, $0<\theta<2$ by some version of Lagrange intermediate value theorem. It is easy to verify that $t^2 w''(t)$ is uniformly bounded for all $x>0$, thus $\alpha_k-2\alpha_{k+1}+\alpha_{k+2}=O(1/k^2)$ as we need.
For even smaller value of $\Re z$ do the same thing, writing something like $\alpha_k=\beta_k+2\beta_{k+1}+\beta_{k+2}$ for $\beta_k=\alpha_k-2\alpha_{k+1}+3\alpha_{k+2}-4\alpha_{k+3}+\dots$.
