$$
\begin{array}{l}
\text { 13. Given the function } f(x)=-2 x+4 \text {, let } \\
S_{n}=f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\cdots+f\left(\frac{n-1}{n}\right)+f(1) \\
\left(n \in \mathbf{N}_{+}\right) \text {. }
\end{array}
$$

If the inequality $\frac{a^{n}}{S_{n}}<\frac{a^{n+1}}{S_{n+1}}$ always holds, find the range of real number $a$.

Three, 13. Since $S_{n}=-2\left(\frac{1}{n}+\cdots+\frac{n-1}{n}+1\right)+4 n=$ $3 n-1$, from $\frac{a^{n}}{S_{n}}0$, then $a^{n}0$, which is a contradiction. Therefore, when $a0$, since $a^{n}>0$, from equation (1),
$$
a>\frac{3 n+2}{3 n-1}=1+\frac{3}{3 n-1}
$$

decreases as $n$ increases.
When $n=1$, $1+\frac{3}{3 n-1}$ reaches its maximum value $\frac{5}{2}$, hence $a>\frac{5}{2}$.