7. If for any positive integer $n \geqslant 3$, we have
$$
\sum_{i=1}^{n} \frac{1}{n+i}+\frac{5}{12} \log _{a}(a-1)>\frac{1}{5},
$$

then the range of real number $a$ is $\qquad$
7. $\left(\frac{1+\sqrt{5}}{2},+\infty\right)$.

Let $f(n)=\sum_{i=1}^{n} \frac{1}{n+i}(n \in \mathbf{Z}, n \geqslant 3)$. Then
$$
\begin{array}{l}
f(n+1)-f(n) \\
=\sum_{i=1}^{n+1} \frac{1}{n+1+i}-\sum_{i=1}^{n} \frac{1}{n+i} \\
=\frac{1}{2 n+1}+\frac{1}{2 n+2}-\frac{1}{n+1} \\
=\frac{1}{2 n+1}-\frac{1}{2 n+2}>0,
\end{array}
$$

i.e., $f(n)$ is strictly increasing.
Thus, for any positive integer $n \geqslant 3$, we have
$$
f(n) \geqslant f(3)=\frac{1}{4}+\frac{1}{5}+\frac{1}{6} \text {. }
$$

Therefore, the range of real numbers $a$ that satisfy
$$
\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{5}{12} \log _{a}(a-1)>\frac{1}{5}
$$

is the range of real numbers $a$ that satisfy $\log _{a}(a-1)>-1$.
Since $a>1$, we have
$$
\begin{array}{l}
\frac{1}{a-1}<1 \\
\Rightarrow a>\frac{1+\sqrt{5}}{2} .
\end{array}
$$

In summary, $a \in\left(\frac{1+\sqrt{5}}{2},+\infty\right)$.