TITLE: Simplify $\sum_{n=0}^{N}\binom{N}{n} \frac{a^{N-n}}{n!} \frac{d^n}{dx^n} f(x)$
QUESTION [6 upvotes]: Simplify the following expression
$$S_N = \sum_{n=0}^{N}\binom{N}{n} \frac{a^{N-n}}{n!} \frac{d^n}{dx^n} f(x), $$
where $a$ is a real number and $f(x)$ is an analytic real function.
What is $\lim_n S_n$? Is there a simple intuitive meaning to it? 
I only found formulas which come close but not quite what I need:
Without the $n!$ in the denominator, the expression can be simpified using the  Leibniz rule
$$ \sum_{n=0}^{N}\binom{N}{n} a^{N-n} \frac{d^n}{dx^n} f(x) =e^{-ax}\frac{d^N}{dx^N}\left[ e^{ax} f(x)\right].$$
Without the binomial coefficient the Taylor series expansion of $f(x)$ around $x$ gives
$$\lim_{n\rightarrow\infty}\sum_{n=0}^{N} \frac{a^{N-n}}{n!} \frac{d^n}{dx^n} f(x) =a^N f\left(x+a^{-1}\right).$$
EDIT:
With an extra set of parentheses
$$\left[\sum_{n=0}^N \binom{N}{n}a^{N-n}\right]\left[\sum_{n=0}^{\infty}\frac{1}{n!} \frac{d^n}{dx^n}f(x)  \right] = (1+a)^N f(x+1) $$
On the other hand
$$
\lim_{z\rightarrow 0}\frac{1}{N!}\frac{d^N}{dz^N}\sum_n A_n z^n \sum_m B_m z^m = 
\lim_{z\rightarrow 0}\frac{1}{N!}\frac{d^N}{dz^N} \sum_{N} \sum_{n} A_{N-n}B_n z^N  = \sum_n A_{N-n}B_n$$
so combining the above two 
$$\sum_n \binom{N}{n} \frac{a^{N-n}}{n!} \frac{d^n}{dx^n} f(x) = \lim_{z\rightarrow 0} \frac{1}{N!}\frac{d^N}{dz^N}\left[ (1+az)^N f(x+z)\right] $$
Is this right? Is there a simpler expression?

REPLY [2 votes]: If your function $f(z) = \sum_{i=0}^\infty a_i z^i,$ then 
$$S_N = \sum_{n=1}^N \binom{N}{n} a^{N-n} a_n.$$ This is a binomial convolution,
So the Exponential Generating function of the sequence $S_N$ equals $\exp(a x) g(z),$ where $g(z)$ is the exponential generating function of of the $a_i.$