Prove that $\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_k}=2$

Question

Let $n$ be a perfect number and $d_1 ,d_2,...d_k$ be the list of all positive divisors of $n$ prove that $$\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_k}=2.$$

I tried to do simplification but it reverses to the original problem.

Answer 1

Hint: multiply by $n$ and assume the divisors are sorted in increasing order. What is $\frac n{d_1}?$