In Bump's book on Automorphic forms, Exercise 2.4.2 asks to prove that, when $G = GL(n, \mathbb R)$ and $f \in C^\infty(G)$, then $$f(ge^X) = \sum_{n=0}^\infty \frac1{n! }(X^n f)(g) \tag{1}$$ for $g \in G$ and $X$ in the Lie algebra of $G$. Here, the action of $X$ on $f$ is defined by $$X f(g) = \frac{d}{dt}f(ge^{tX})\vert_{t=0}$$
I am worried about the convergence of the series in the RHS. If the series $$\sum_{n=0}^\infty \frac{(1+\epsilon)^n}{n! }(X^n f)(g) \tag{1}$$ converges for some $\epsilon >0$, then I agree with $(1)$ (it is just the ordinary Taylor expansion of a function of one real variable.) But if $f$ is supported on some ball, $g$ is on the boundary of that ball and $f(ge^X) \neq 0$, then it seems to me that the series in the RHS is $0$ while the LHS is not. (Take for example $n=g=X=1$, so that $Xf(g) = f'(g)$.)
Is $(1)$ actually true?