Solution
= Solution
The expression in the PDF is $1/a^x-1$, with the subtraction outside the fraction. Using the <exponential series> with $a^{-x}=e^{-x\ln a}$ gives
$$
\boxed{a^{-x}-1=-(\ln a)x+\frac{(\ln a)^2}{2}x^2+O(x^3).}
$$
For $a>0$, $a\ne1$, these are the first two nonzero terms. If $a=1$, the <function> is identically zero, so there are no nonvanishing terms.