I am trying to prove the following theorem:
Theorem. A number is perfect iff the sum of the reciprocal of its divisors, excluding $1$, is $1$.
Thus far, this is the proof that I have managed to sew:
Proof. Let $n$ be perfect. Then $2n=1+a_1+a_2+\cdots+a_m+n$, where each $a_j$ is a divisor of $n$. Moreover, let $1. It follows that $2=\frac{1}{n}+\frac{a_1}{n}+\frac{a_2}{n}+\cdots+\frac{a_m}{n}+1\Longrightarrow$ $1=\frac{1}{n}+\frac{1}{a_m}+\frac{1}{a_{m-1}}+\cdots+\frac{1}{a_1}$, as required. Conversely, let $1=\frac{1}{n}+\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_m}$. It follows that $1=\frac{1}{n}+\frac{a_m}{n}+\frac{a_{m-1}}{n}+\cdots+\frac{a_1}{n}\Longrightarrow$ $n=1+a_1+a_2+\cdots+a_m$. Therefore, $n$ is perfect. $\square$
Nevertheless, I do not feel very confident about it. What do you guys think?
elementary-number-theory
perfect-numbers