Why is the sum of the reciprocals of all of the divisors of a perfect number is equal to 2?

Answer 1

Suppose the perfect number P has divisors f1, f2, ... , fk, P (where all the fs are smaller than P.

Then, by the definition of a perfect number, f1 + f2 + ... + fk = P

So that f1 + f2 + ... + fk + P = 2P

Dividing through by P,

f1/P + f2/P + ... + fk/P + P/P = 2 . . . . . . . . . . . (2)

Now, since f1 is a factor of P, the f1*g1 = P where g is the cofactor and so f1/P = 1/g1.

That is, f1/P is the reciprocal of one of the other factors of P. Also, the last term on the left is the reciprocal of the factor 1.

And therefore,

f1/P + f2/P + ... + fk/P + P/P = 2 = 1/g1 + 1/g2 + ... + 1/gk + 1/1