Let $X_n$ = $\{1,2,3,4,\ldots,n\}$ (a set).
Conjecture and prove that $\sum_{\emptyset \neq A\subseteq X_n}\frac{1}{p_A}=n$, where $p_A$ is the product of the subset.
Attempt: $\sum_{\emptyset \neq A\subseteq X_n}\frac{1}{p_A}=n$. Shown true for n=1,2,3,4.(omitted)
$\sum_{\emptyset \neq A\subseteq X_n}\frac{1}{p_A}=n$ true.
$\sum_{\emptyset \neq A\subseteq X_n \cup X_{n+1}}\frac{1}{p_A}=n$ $+\sum_{\emptyset \neq A \wedge A\nsubseteq X_n \wedge A\subseteq X_{n+1}}\frac{1}{p_A}=n+1$
Thanks.
Note that every subset of $X_{n+1}$ is either a subset of $X_n$, or a subset of $X_n\cup\{n+1\}$.
Therefore we can re-write the sum: $$\begin{align} \sum_{\varnothing\neq A\subseteq X_{n+1}}\frac1{p_A}&=\sum_{\varnothing\neq A\subseteq X_n}\frac1{p_A}+\frac1{n+1}+\sum_{\varnothing\neq A\subseteq X_n}\frac1{(n+1)p_A}\\ &=\frac1{n+1}+\left(1+\frac1{n+1}\right)\sum_{\varnothing\neq A\subseteq X_n}\frac1{p_A}\\ &=\frac1{n+1}+\left(1+\frac1{n+1}\right)n\\ &=n+\frac n{n+1}+\frac1{n+1}=n+1 \end{align}$$