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)



Strange limit problem involving $\binom{z}{n} e^{-xn\log n}$ with $z \in \mathbb{C}$
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 $\beta\_k=\alpha\_k-2\alpha\_{k+1}+3\alpha\_{k+2}-4\alpha\_{k+3}+\dots$.


