Suppose that there are k divisors of n: such that
Then,
Divide both sides by n and you get ...
Let n in Z with n>0. Prove that the sum of 1/d = sigma(n)/n.
d is the divisors n. (d|n,d>o)
I must be doing something wrong but i need help spotting it.
For the sum of (1/d) I sub in 1/p1^a1+p2^a2+...Pr^ar.
Then for sigma(n) I have
[(p1^(a1+1) - 1)/P1-1] [(P2^(a2+1) - 1)/ P2-1]...[Pr^(ar+1)-1/pr-1]
and for n i used an arbitrary prime factorization. n=p1^a1P2^a2...pr^ar
I am not sure if started off right. but my calculations does not prove that statement.
Did i start the wrong way and if i did not how can I can i compute to prove the statement.