From Theorem 2 in § 2, we know
$$\sum_{d \mid a} \frac{1}{d}=\sum_{d \mid a} \frac{1}{(a / d)}=\frac{1}{a} \sum_{d \mid a} d=\frac{1}{a} \sigma(a) .$$

From this and Example 2, we can obtain
$$\sum_{d \mid 180} \frac{1}{d}=\frac{1}{180} \sigma(180)=\frac{91}{30} .$$