Solution
= Solution
Use the <Maclaurin series> $\ln(1+x)=x-x^2/2+x^3/3+O(x^4)$ and the <geometric series> $(1+y)^{-1}=1-y+y^2-y^3+\cdots$. Multiplication, grouped by total degree, gives
$$
\boxed{\frac{\ln(1+x)}{1+y}=x-\frac{x^2}{2}-xy+\frac{x^3}{3}+\frac{x^2y}{2}+xy^2+O((|x|+|y|)^4).}
$$
The six displayed nonzero terms include every contribution through total degree three. A sufficient open region for the convergent product expansion is $|x|<1$, $|y|<1$.