If $S(n)$ is an odd integer, what is the sum of all possible $\frac1n?$ If $n$  is a positive integer, let $S(n)$ be the sum of all the positive divisors of $n$.
If $S(n)$ is an odd integer, what is the sum of all possible $\frac1n?$
The function $S$ is multiplicative and so, if we have the prime factorisation $n = p_1^{a(1)}p_2^{a(2)} \cdots p_m^{a(m)}$, where $p_1,p_2,...,p_m$ are distinct primes, then how will I continue to solve this?
 A: 
The function $S$ is multiplicative and so, if we have the
  prime factorisation $n = p_1^{a(1)}p_2^{a(2)} \cdots
> p_m^{a(m)}$, where $p_1,p_2,...,p_m$ are distinct primes.

Continuing what I said
$ S(n) \; = \; \prod_{k=1}^m S\big(p_k^{a(k)}\big) \; = \; \prod_{k=1}^m \big(1 + p_k + p_k^2 + \cdots + p_k^{a(k)}\big) $
Note that $S(2^n) = 2^{n+1}-1$ is odd for all $n \ge 1$, while $S(p^n)$ is odd precisely when $n$ is even for any odd prime $p$.
If $S(n)$ is to be odd, we must have $S\big(p_k^{a(k)}\big)$ odd for all $k$, and hence the index of any odd prime factor of $n$ must be even. Thus the set of numbers $n$ for which $S(n)$ is odd is the set of perfect squares, together with the set of twice the perfect squares. Thus we need to evaluate
$ \sum_{n=1}^\infty \frac{1}{n^2}+ \sum_{n=1}^\infty \frac{1}{2n^2} \; = \; \tfrac32\zeta(2) \; = \; \boxed{\tfrac14\pi^2} $
A: Hint The divisors of $p_i^{a(i)}$ are $1, p_i, p_i^2,...,p_i^{a(i)}$. Their sum is.....
A: $S(n)$ is multiplicative, so $S(n) = \prod_{i=1}^m S(p_i^{a_i})$. If any of the $S(p_i^{a_i})$ are even, the whole product will be even. So what is the condition on $S(p^a)$ being even for some prime $p$ and natural number $a$? What does that tell you about the set of numbers $n$ for which $S(n)$ is odd?
Once you've figured this out, you're going to need to sum the series. You'll want to factor out a geometric series involving powers of $2$, and you'll need to use the fact that $\sum_{k=0}^\infty k^{-2}  = \pi^2/6$.
