Approximating the summation of the logarithm of a function I am working through a textbook for self-study and came across an approximation that I'm unsure how to solve
$$
\sum_{n=1}^{N} \mathrm{ln}[A - (n-1)\epsilon ] = N \mathrm{ln}[A - (N-1)\epsilon/2] + \mathcal{O}(\epsilon^{2}),
$$
where $A$ and $\epsilon$ are positive constants. We can assume $N\epsilon \ll A$. I tried using the Euler-Maclaurin sum formula but I can't get the approximation back. I'm starting to think there's something simple that I missed, perhaps with a geometric summation. Thoughts?

Here's some naive fiddling.
Writing $c$ for $\epsilon$
because lazy,
using $=\ln(1-c)=-c+O(c^2)$,
and assuming that $c \to 0$,
$\begin{array}\\
\sum_{n=1}^{N} \ln[A - (n-1)c ] 
&=\sum_{n=1}^{N} (\ln(A)+\ln[1 - (n-1)c/A ]\\ 
&=N\ln(A)+\sum_{n=1}^{N} \ln(1 - (n-1)c/A)\\ 
&=N\ln(A)-\sum_{n=1}^{N} ((n-1)c/A+O(c^2)))\\ 
&=N\ln(A)-N(N-1)c/(2A)+O(Nc^2)
\qquad\text{assuming the big-oh is uniform}\\ 
&=N(\ln(A)-(N-1)c/(2A))+O(Nc^2)\\
&\approx N(\ln(A)+\ln(1-(N-1)c/(2A))+O(Nc^2)\\
&= N(\ln(A(1-(N-1)c/(2A)))+O(Nc^2)\\
&= N(\ln(A-(N-1)c/2))+O(Nc^2)\\
\end{array}
$
