23 Given the inequality $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}>\frac{1}{12} \cdot \log _{a}(a-1)+\frac{2}{3}$ holds for all natural numbers $n$ greater than 1, find the range of real values for $a$.

For all natural numbers $n > 1$, the inequality $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}>\frac{1}{12} \cdot \log _{a}(a-1)+\frac{2}{3}$ is always true. Determine the range of real values for $a$.
23 Let $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}$. Since
$$
f(n+1)-f(n)=\frac{1}{2 n+1}+\frac{1}{2 n+2}-\frac{1}{n+1}=\frac{1}{2 n+1}-\frac{1}{2 n+2}>0 .
$$

Therefore, $f(n)$ is increasing. We only need $f(2)>\frac{1}{12} \log _{a}(a-1)+\frac{2}{3}$, i.e.,
$$
\frac{1}{3}+\frac{1}{4}>\frac{1}{12} \log _{a}(a-1)+\frac{2}{3}
$$

Thus,
$$
-1>\log _{a}(a-1)
$$

Hence, $a>1$ and
$$
\frac{1}{a}>a-1
$$

Solving this, we get $1<a<\frac{\sqrt{5}+1}{2}$.