Sum of inverse product of subsets 
Define $I_s$ as the inverse of the product of the elements in $S$. If $S=\{1,2,3\}$, then $I_s=\frac{1}{1*2*3}$. What is the sum of all $I_s$ if $S$ is a nonempty subset of the first N positive integers?

I'm pretty sure the sum is always N. If N=4,
$$\Sigma({I_s})= (1+1/2+1/3+1/4)+(1/2+1/3+1/4+1/6+1/8+1/12)+(1/24+1/12+1/8+1/6)+1/24$$
Each fraction is repeated two times except 1, since $I_s=\frac{1}{a*b*c...}=\frac{1}{1*a*b*c...}$.
Thanks!
 A: Take any $n \leq N$. $\frac{1}{N}$ is a term in exactly half of all subsets of $S$. If the other terms have some sum $\sigma$, the total sum is $\sigma \left( 1 + \frac{1}{n}\right)$. Factor out that $\frac{n+1}{n}$ for every $n$ in $S$ until you get the trivial product 1, and this is still equivalent to your sum. Thus, multiplying all these terms together one gets $$ \frac{N+1}{N }\cdot \frac{N}{N-1} \cdot ...\cdot \frac32 \cdot \frac21 = n+1 $$ and, as you've noted this doubly counts by adding in the empty set as a subset with product 1. So the sum is, as you say, $N$
A: When $n$ is odd.  Let $p(x)$ be a polinomial with $p(x)[x^n]=1$ and $\frac{1}{i}\quad\forall i\in 1,...,n$ its roots. In this case, $ p(-1) -1= \Sigma(I_N) $ due to the Girard relations. Try to build this polinomial using 
$$(-1) \prod_{i=1}^{n} (x-\frac{1}{i})$$
If $n$ is even
$$ \prod_{i=1}^{n} (x-\frac{1}{i})$$
This would be my first attempt. I am not sure!!
Update 
The root $1$ appears once, then it is needed to remove $n-1$ from this product. It is esier to use $$\prod_{i=2}^{n} (1+\frac{1}{i})$$
