# The sum of postive divisors Function

• August 4th 2008, 04:49 AM
JCIR
The sum of postive divisors Function
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.
• August 4th 2008, 10:24 AM
o_O
Suppose that there are k divisors of n: $d_{1}, d_{2}, \ldots, d_{k}$ such that $d_{1}d_{k} = d_{2}d_{k-1} = ... = n$

Then, $d_{k} + d_{k-1} + \ldots + d_{2} + d_{1} = \sigma (n)$

Divide both sides by n and you get ...