Finite exponential power

Function

 context $m\in{\mathbb N} definition${\rm pexp}_n: \mathbb C\to\mathbb C$definition${\rm pexp}_n(z) := \left(1 + \dfrac {x} {n} \right)^n {\rm pexp}_n(x) = \sum_{k=0}^n a_k(n)\dfrac {1} {k!} x^k $with$a_k(n)=\prod_{j=1}^{k-1}\left(1-\dfrac{k-j}{n}\right)\le 1$Elaboration$\left(x+y\right)^m=\sum_{k=0}^m \dfrac{n!}{k!\,(m-k)!} x^k y^{m-k}$so$\left(1 + b(n)\,x \right)^n = \sum_{k=0}^n \left( b(n)^{-k}\dfrac {n!} {(n-k)!} \right) \dfrac {x^k} {k!}$(Note that here the summands depend on the upper sum bound$n$, this sum doesn't make for an infinite sum of partial sums - the to be partial sums are all different) The above sum follows. Also,$= \sum_{k=0}^n \left(\prod_{j=1}^{k}\left(\dfrac{1}{j}-\dfrac{1}{n}\left(\frac{k}{j}-1\right)\right) x \right)$Derivative$\frac{\mathrm d}{\mathrm d z}{\rm pexp}_n(z)=\dfrac{1}{1+z/n}{\rm pexp}_n(z)\frac{\mathrm d}{\mathrm d n}{\rm pexp}_n(z)=-\left(\dfrac{1}{1+(z/n)^{-1}}+\log\left(\dfrac{1}{1+z/n}\right)\right){\rm pexp}_n(z)\$