Sum of the reciprocals of divisors of a perfect number is $2$?

Question

How do I show that the sum of the reciprocals of divisors of a perfect number is $2$?

I tried $d_i\mid n$ with $i\in\mathbb{N},\;d_i\leq n$ then $$\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}+...+\frac{1}{d_i}=2$$ $$\sum_{d\mid n} \frac{1}{d}=2$$ So actually, I have to show this last equality, whereas $n$ is a perfect number.

Answer 1

Hint: multiply the first equation by $n$, then note that if the divisors are sorted in increasing order $\frac n{d_1}=?$

Answer 2

Clear the denominators (assume that $n$ is not a perfect square). The denominator is $n^{d(n)/2)}.$ Each term in the numerator is $n^{d(n)/2-1} d_k,$ for some divisor $d_k$ of $n.$ So, the sum is equal to $$\frac{1}{n} \sum_{d|n|} d = 2.$$