I am stuck with trying to understand a Taylor expansion in an article I'm reading. The author claims that
Here, $\xi=X_{i+1}-X_i$ is the time increment of a random walk and $o(\tau)$ collects all terms that go to zero faster than $\tau$. In particular, I can't see how the $e^{\rho \tau}$ cancels. Later in the text, the author claims that $(e^{\rho \tau}-1)F(X_i)$ expands to $\rho \tau F(X_i)+o(\tau)$. Any ideas on how can I obtain this result?
"In particular, I can't see how the $\mathrm{e}^{\rho \tau}$ cancels."
It doesn't. Its first term is $1$ and that is the only term that survives when everything else is swept into the little-o. $$ \mathrm{e}^{\rho \tau} = 1+\rho\tau + \frac{(\rho\tau)^2}{2}+ \frac{(\rho\tau)^3}{6} + \cdots $$ so, assuming $f(X_i)$ is not hiding a $\tau$-dependence, $$ \mathrm{e}^{\rho \tau} f(X_i) \tau = f(X_i)\tau + f(X_i)\rho \tau^2 + \frac{f(X_i) \rho^2 \tau^3}{2} + \cdots $$ Here, you are interested in the asymptotics as $\tau \rightarrow 0$, so to match the usual definition of little-o, you should make the change of variable $\tau \mapsto 1/t$ and look at what happens as $t \rightarrow \infty$. In this case, $f(X_i)\rho (1/t^2)$ goes to $0$ faster than $1/t$, and similarly for the terms with higher powers of $\tau$. So only the first term is not $o(\tau)$ and the rest are swept into $o(\tau)$.
Your later result is the same thing. The "${}-1$" removes the leading $1$ from the series for the exponential. The first remaining term is $\rho \tau F(X_i)$. As in the first case, this is the only term with linear dependence on $\tau$, so is not swept into $o(\tau)$ and all other terms are swept into the little-o.