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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.

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

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  • $\begingroup$ $$n\sum_{d\mid n}\left(\frac{1}{d} \right)=\sum_{d\mid n} \frac{n}d=s(n)$$ $\endgroup$ Dec 5 '13 at 17:00
  • $\begingroup$ Ahhhhhhhhhhhhh.... $$s(n)=2n\Longrightarrow \sum_{d\mid n} \frac{n}{d}=2n\Longrightarrow \sum_{d\mid n}\frac{1}{d}=2$$ Correct? $\endgroup$ Dec 5 '13 at 17:03
  • $\begingroup$ That is correct. You lost the subscript on $d$ in these comments, but you have the point. $\endgroup$ Dec 5 '13 at 17:04
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$$\displaystyle \sum_{d \mathop \backslash n} d = \sigma \left({n}\right) \leadsto\dfrac 1 n \sum_{d \mathop \backslash n}=\dfrac {\sigma \left({n}\right)} n\leadsto \sum_{d \mathop \backslash n} \frac d n = \dfrac {\sigma \left({n}\right)} n \leadsto \sum_{d \mathop \backslash n} \frac 1 d =\dfrac {\sigma \left({n}\right)} n=2 $$

source: proofwiki

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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 $$\frac{1}{n} \sum_{d|n|} d = 2.$$

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