TITLE: Closed form/ meaning of sum of geometric series of operator exponentials
QUESTION [3 upvotes]: I'm taking my first class on quantum mechanics right now and we've been using various operators throughout it; one operator we derived was the displacement operator, $$e^{a\frac{d}{dx}}f(x)=f(x+a)$$ I was playing around with Riemann sums and the previous identity, and I "derived" a strange formula through analogy:
$$\frac{a}{n}\sum_{i=0}^n{f\left(x+a\frac{i}{n}\right)}=\frac{a}{n}\left[\sum_{i=0}^n{\left(e^{\left(\frac an\right)D}\right)^i}\right]f(x)=\frac{a}{n}\cdot\frac{1-e^{a\frac{n+1}{n}D}}{1-e^{\frac{a}{n}D}}f(x) \longrightarrow \frac{e^{aD}-1}{D}f(x)$$
With the last identity coming from making $n$ very large. I'm wondering if this formula has any merit, or at least if the geometric sum of operator exponentials has any actually closed. Additionally, is there a field of study which deals with these types of formulas?
I dont think this approach is really valid since $\frac{1}{\text{operator}}$ doesn't seem to be anything meaningful innately. But, If $\frac{1}{D} (f)$ is taken to mean computing the antiderivative of $f(x)$, it seems this formula actually would make sense.

REPLY [1 votes]: Yes, formal operator expressions such as 
$$\frac{e^{aD}-1}{D} \sim a\sum_{n=0}^\infty \frac {(aD)^n}{(n+1)!}$$
are understood to represent their power series, "with all suitable conditions satisfied where they must", informally now. 
You note you never had to consider the inverse of D. 
A very similar expression for commutators, the adjoint endomorphism, ad$_Y ~X\equiv [Y,X]$, for operators X,Y, which behaves like a derivative, is ubiquitous in  the CBH expansion, a cornerstone of Lie group theory,
$$
\log(\exp X\exp Y) = X + \left ( \int^1_0 \psi \left ( e^{\operatorname{ad} _X} ~ e^{t \,\text{ad} _ Y}\right ) \, dt \right) \, Y, 
$$
where 
$$
 \psi(e^x)= \frac{x}{1-e^{-x}} \\
=\sum_{n=0}^\infty B_n ~ \frac{x^n}{n!}~, $$
the generating function of the Bernoulli numbers, B0 = 1, B1 = 1/2, B2 = 1/6, B4 = −1/30, ...
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