Example 2 (2000 Hebei Province Competition Question) When $a>1$, if the inequality $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}>\frac{7}{12} \cdot\left(\log _{a+1} x-\right.$ $\left.\log _{a} x+1\right)$ holds for all positive integers $n$ not less than 2, then the range of $x$ is ( ).
A. $21$
Solution: Choose D. Reason: Simplify the original inequality to a simple inequality form. Let $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}$, then $f(n+1)=f(n)+\frac{1}{2 n+1}+\frac{1}{2 n+2}-\frac{1}{n+1}=f(n)+\frac{1}{2 n+1}-\frac{1}{2 n+2}>f(n)$.
Therefore, $f(n)$ is an increasing function, so $f(n) \geqslant f(2)=\frac{7}{12}$.
Given that the inequality always holds, we have
$\frac{7}{12}\left(\log _{a+1} x-\log _{a} x+1\right)>1$, thus $x>1$.