4. Given that $n$ is a positive integer, if $\frac{n^{2}+3 n-10}{n^{2}+6 n-16}$ is a reduced fraction, then the value of this fraction is $\qquad$ .
Consider $\frac{8}{11}$.
Analyzing $\frac{n^{2}+3 n-10}{n^{2}+6 n-16}=\frac{(n+5)(n-2)}{(n+8)(n-2)}$, when $n-2= \pm 1$, if $(n+8, n+5)=(n+5,3)=1$, then $\frac{n^{2}+3 n-10}{n^{2}+6 n-16}$ is a reduced fraction. Therefore, when $n=3$, the fraction is a reduced fraction, so this fraction is $\frac{8}{11}$.