smc

The error made in this approximation is $\frac{1}{1+y}-(1-y)$, and the correct value is $\frac{1}{1+y}$. Taking the ratio of these two values, we have: $$\begin{array}{rl}& \\ \frac{\frac{1}{1+y}-(1-y)}{\frac{1}{1+y}}& =\frac{[\frac{1}{1+y}-(1-y)][1+y]}{[\frac{1}{1+y}][1+y]}\\ & =\frac{1-(1-y^2)}{1}\\ & =y^2\end{array}$$ Thus, our answer is $\boxed{y^2}$