We are given a set of natural numbers 2, 3, 4, …, n.
Consider all subsets, each of them consisting of the combinations $_{(n-1)}C_{2}$, $_{(n-1)}C_{3}$, $_{(n-1)}C_{4}$ etc. We take the products of the terms in each such subset and then its reciprocals.
Find the sum of all these reciprocals.
I believe I must find a recursive formula but I have no idea how to proceed…
I tried for n=4 and n=5 and found 5/12 and 86/120 respectively but I don't know how to continue.
For example, for n=4: Let's consider the set {2,3,4}. Then we have the subsets {2,3}, (2,4), (3,4), (2,3,4) and their respective elements' products are 6, 8, 12 and 24. Then we take $\frac{1}{6}, \frac{1}{8}, \frac{1}{12}, \frac{1}{24}$ and add them. The result is $\frac{10}{24}$.
Also noticed that for each such fraction, the product of the numerator and the denominator equals n!.
Best Answer
Let $x_n$ denote your sum of the reciprocals of all products of subsets of $\{ 2, \ldots, n \}$ with at least 2 elements. Then $$ x_n + \left( \frac 12 + \ldots + \frac 1n \right) $$ is the sum of the reciprocals of all products of non-empty subsets of $\{ 2, \ldots, n \}$, and $$ \frac 11 + 2 x_n + 2 \left( \frac 12 + \ldots + \frac 1n \right) $$ is the sum of the reciprocals of all products of non-empty subsets of $\{ 1, \ldots, n \}$. According to Show that the sum of reciprocal products equals $n$ that sum is exactly $n$, therefore $$ x_n = \frac 12 (n-1) - \left( \frac 12 + \ldots + \frac 1n \right) = \frac 12 (n+1) - H_n $$ where $H_n$ is the $n^\text{th}$ harmonic number.