How perfect are the reciprocals of the factors??

We know that the divisors come in pairs, ie. $d_{i} \times d_{j} = n$ for some $i,j$ . Therefore, WLOG, say $d_{2i-1} \times d_{2i} = n$ for all $1 \le i \le \dfrac{k}{2}$ . Then we can combine the fractions of each pair, meaning the statement becomes: $\dfrac{d_{1}+d_{2}}{d_{1} \times d_{2}} + \dfrac{d_{3}+d_{4}}{d_{3} \times d_{4}} + ... + \dfrac{d_{k-1}+d_{k}}{d_{k-1} \times d_{k}}$ But the denominator of all these fractions is just $n$ , so we can group all the fractions, meaning the statement becomes: $\displaystyle\sum_{i=1}^{k} \dfrac{d_{i}}{n}$ We know that $n$ is a perfect number, so $\displaystyle\sum_{i=1}^{k} = 2n$ . Therefore the statement becomes: $\dfrac{2n}{n} = \fbox{2}$

take an example. for 6, divisors are 1,2,3 and 6. so 1/1+1/2+1/3+1/6=2