| 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
$\frac{\mathrm d}{\mathrm d z}{\rm pexp}_n(z)=\dfrac{1}{1+z/n}{\rm pexp}_n(z)$
$\frac{\mathrm d}{\mathrm d n}{\rm pexp}_n(z)=-\left(\dfrac{1}{1+(z/n)^{-1}}+\log\left(\dfrac{1}{1+z/n}\right)\right){\rm pexp}_n(z)$
Wikipedia: Exponential function, Matrix exponential, Exponential map