Perfect Numbers and their reciprocals.

[Slats18]

For those who don't know, here's the definition:

A Perfect Number is some number n such that the sum of its divisors is equal to 2n (or just n if you don't count n|n =P )
Ex: Let n=6, and the divisors of 6 are 1,2,3 and 6

1+2+3+6 = 12 = 2*6.

Moving on, there's also the neat fact about them that states that the reciprocals of the divisors will always add up to 2.

Ex: n=6, 1/6 + 1/3 + 1/2 + 1/1 = 2

Now, my question/wonderment: is there any possible relation between this fact, and the summation of (1/2)^n, which converges to two as n approaches infinity?

 

[Dickfore]

Let

$$1 < p_{1} < p_{2} < \ldots < p_{2m} < N$$

be all the divisors of $N$ excluding 1 and itself.

Notice that the number of such divisors must be even and that:

$$p_{1} p_{2 m} = p_{2} p_{2 m - 1} = \ldots p_{m - 1} p_{m + 1} = N$$

If $N$ is perfect then we also have:

$$N = 1 + p_{1} + \ldots + p_{2m}$$

Divide this equality by $N$ and you get:

$$1 = \frac{1}{N} + \frac{p_{1}}{N} + \ldots + \frac{p_{2m}}{N}$$

But, according to the upper equalities:

$$\frac{p_{1}}{N} = \frac{1}{p_{2m}}, \frac{p_{2m}}{N} = \frac{1}{p_{1}}, \ldots$$

and we get:

$$1 = \frac{1}{N} + \frac{1}{p_{2m}} + \ldots + \frac{1}{p_{1}}$$

Q.E.D.

 

[Slats18]

Thank you for the speedy reply, Dickfore, and thank you on the proof of why the fact holds. But, I'm more interested in the discussion of whether they are related in any way (excluding the trivial), not just proof-wise =)