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Suppose N is a perfect number. Then N cannot be a square number and so N has an even number of factors.Suppose the factors are f(1) =1, f(2), f(3), ... , f(k-1), f(k)=N.

Furthermore f(r) * f(k+1-r) = N for r = 1, 2, ... k so that f(r) = N/f(k+1-r)

which implies that 1/f(r) = f(k+1-r)/N


Then 1/f(1) + 1/(f(2) + ... + 1/f(k)

= f(k)/N + f(k-1)/N + ... + f(1)/N

= [f(k) + f(k-1) + ... + f(1)] / N

= 2N/N since, by definition, [f(k) + f(k-1) + ... + f(1)] = 2N

Q: Why is the sum of the reciprocals of all of the divisors of a perfect number equal to 2?
This answer is:
Related answers

Suppose N is a perfect number. Then N cannot be a square number and so N has an even number of factors.Suppose the factors are f(1) =1, f(2), f(3), ... , f(k-1), f(k)=N.

Furthermore f(r) * f(k+1-r) = N for r = 1, 2, ... k so that f(r) = N/f(k+1-r)

which implies that 1/f(r) = f(k+1-r)/N


Then 1/f(1) + 1/(f(2) + ... + 1/f(k)

= f(k)/N + f(k-1)/N + ... + f(1)/N

= [f(k) + f(k-1) + ... + f(1)] / N

= 2N/N since, by definition, [f(k) + f(k-1) + ... + f(1)] = 2N

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