The sum is $2$

Let $n$ be a perfect number. Prove that:

$$\mathbf{\sum_{d \mid n} \frac{1}{d}= 2} $$

Solution

We recall that a perfect number is a positive integer that is equal to the sum of its proper positive divisors, that is, the sum of its positive divisors excluding the number itself. Equivelanlty we have that:

$$\sigma(n)=2n$$

Thus for the sum in question we have that:

$$\mathbf{\sum_{d \mid n} \frac{1}{d}= \frac{\sigma(n)}{n}= \frac{2n}{n}=2} $$

ending the exercise.