A expression $a(x,h) \in \mathbb{C}(h)[x]^\mathrm{alg}$ imply that $$ a(x,h)^n+p_{n-1}(h)[x]a(x,h)^{n-1}+\cdots+p_0(h)[x]=0 $$
For example $e^{dh}x \notin \mathbb{C}(h)[x]^\mathrm{alg}.$ Where $d$ could be treated as constant.
I want to show the following expression does not belong $\mathbb{C}(h)[x]^\mathrm{alg}$. We treat $d$ as a constant. $$ \sum_{k=1}^{d}(-1)^k\binom{d-1}{k-1}e^{d(k-\frac{d+1}{2})h} \notin \mathbb{C}(h)[x]^\mathrm{alg}. $$ It has an exponential in the expression and looks like the example but I still could not conclude. Any help appreciated.
I was wondering about the following way to answer my question but I think I am bit sloppy. Lets denote the expression by $$f(h):=\sum_{k=1}^{d}(-1)^k\binom{d-1}{k-1}e^{d(k-\frac{d+1}{2})h} \notin \mathbb{C}(h)[x]^\mathrm{alg}. $$ Now as $\lim_{h\rightarrow \infty} f(h)\rightarrow \pm \infty.$ depending on $d$ being odd or even. Not only that I guess it's asymtoptically equivalent to $e^{dh}$ hence $f(h)\notin \mathbb{C}(h)[x]^\mathrm{alg}.$