Question. Is anything interesting known about the "Dedekind constant" $$\sum_{n = 0}^\infty \frac{1}{M(n)},$$ where $M(n)$ is the $n$th Dedekind number?
Motivation. Given a sequence of positive real numbers $s$, define $\langle s \rangle$ to be the sum of reciprocals of the elements of $s$. $$\langle s \rangle := \sum_{n = 0}^\infty \frac{1}{s_n}$$ It's best to take the limit in $[0,\infty]$ so that sequences that grow too slowly have $\langle s \rangle$ equal to $\infty$. Wikipedia has a page about the values of $\langle \Box \rangle$ for certain sequences. Some familiar examples include $$\langle n \mapsto 2^n \rangle = 2, \qquad \langle n \mapsto n! \rangle = e, \qquad \zeta(s) = \langle n \mapsto (n+1)^s \rangle$$
Now, here's something interesting about the $2^n$ example: If $X$ is a finite set, then the bounded semilattice $FX$ freely generated by $X$ is isomorphic to the powerset lattice $(\mathcal{P}(X), \cup, \emptyset).$ Thus if $X$ has $n$ elements, then $FX$ has $2^n$ elements. So the sequence $n \mapsto 2^n$ arises from the concept "bounded semilattice" in a natural way.
Other choices of algebraic structure can also be used. However, one must be careful, because free algebras often have infinitely many elements. For example, the abelian group freely generated by $1$ or more elements is infinite. However, in some cases we end up with finite numbers. For example the concept "bounded distributive semilattice" has this property, and gives rise to the Dedekind numbers in the same way. It seems possible that the associated value of $\langle \Box \rangle$ is significant in some way or another.