Who cares about perfection?

Suppose a perfect number $n$ has factors $\frac{n}{a_1}, \frac{n}{a_2}, \frac{n}{a_3},....., \frac{n}{a_k}$ .

So, the divisors would be $1, \frac{n}{a_1}, \frac{n}{a_2}, \frac{n}{a_3},....., \frac{n}{a_k}, n$

Hence the sum of their reciprocals would be $1+\frac{1}{\frac{n}{a_1}}+\frac{1}{\frac{n}{a_2}}+.....+\frac{1}{\frac{n}{a_k}}+\frac{1}{n}\\\rightarrow 1+\frac{a_1}{n}+\frac{a_2}{n}+....+\frac{a_k}{n}+\frac{1}{n}\\\rightarrow 1+\frac{a_1+a_2+a_3+....+a_k+1}{n}$

Since $a_1, a_2, a_3 ....a_k$ divide $n$ it means they themselves are factors.

We know that perfect numbers are the sum of their divisors excluding the number it self. This implies that $1+a_1+a_2+a_3+....+a_k=n$

Hence $1+\frac{a_1+a_2+a_3+....+a_k+1}{n}\\=1+\frac{n}{n}\\=1+1=\huge{\boxed{2}}$

For a perfect number $1+d_1+d_2+...+d_{k}+n=2n$ divide by $n$ $\frac{1}{n}+\frac{d_1}{n}+\frac{d_2}{n}+...+\frac{d_k}{n}+1=2 \\ \frac{1}{n}+\frac{1}{d_k}+\frac{1}{d_{k-1}}+...+\frac{1}{d_1}+1=2$