A summation involving the inverses of the products of the elements of subsets of a set

Question

A big thank you in advance to all who have sacrificed their time to help me with the following problem.

Consider a set $T$ containing the first $k$ natural numbers. First, we find all the $v$-combinations of $T$, where $1\leq v\leq k$. There are $\binom{k}{v}$ such subsets. Now find the product $P_i$ of the elements of each of these $v$-combinations. Take $\frac{1}{P_i}$. Is there a formula to determine the summation over all $\frac{1}{P_i}$ of $T$?

An example might help: suppose $T=\{1, 2, 3, 4\}$. Then all the $3$-combinations of $T$ are $$T_1=\{1, 2, 3\}, T_2=\{1, 2, 4\}, T_3=\{1, 3, 4\}, T_4=\{2, 3, 4\}.$$ The products of the elements of the various subsets are: $$P_T{_1}=6, P_T{_2}=8, P_T{_3}=12, P_T{_4}=24.$$ We then take the sum of the reciprocals of these products: $$1/6+1/8+1/12+1/24=5/12.$$

So the question is whether there is a known formula for summing the reciprocals of the products of the elements of all $v$-combinations of a set of the first $k$ natural numbers.

Thank you once again!!

Answer 1

We have using generating functions the closed form

$$[x^v] \prod_{q=1}^k \left(1+\frac{x}{q}\right) = \frac{1}{k!} [x^v] \prod_{q=1}^k (x+q) \\ = \frac{1}{k!} [x^{v+1}] \prod_{q=0}^k (x+q) = \frac{1}{k!} {k+1\brack v+1},$$

where we have used Stirling numbers of the first kind.