Is the sum of reciprocals of all products from $2$ to $n-1$ always $0.5n-1$? I was looking up riddles for my math classes to work on for the end of the year and found the following riddle. http://mathriddles.williams.edu/?p=129
I followed the advice and started working with examples of small numbers and stumbled upon a pattern that I wanted to generalize.
$$\frac{1}{2}(1)=0.5$$
$$\frac{1}{3}\left(1+\frac{1}{2}\right)=0.5$$
$$\frac{1}{4}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{2\cdot 3}\right)=0.5$$
$$\frac{1}{5}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{2\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 4}+\frac{1}{2\cdot 3\cdot 4}\right)=0.5$$
$$\frac{1}{6}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{2\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{2\cdot 5}+\frac{1}{3\cdot 4}+\frac{1}{3\cdot 5}+\frac{1}{4\cdot 5}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{2\cdot 3\cdot 5}+\frac{1}{2\cdot 4\cdot 5}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{2\cdot 3\cdot 4\cdot 5}\right)=0.5$$
If my pattern doesn't make sense, I'm taking $\frac{1}{n}$ and multiplying it by the sum of the reciprocals of all unique products for $2$ to $n-1$ and it comes out to 0.5 each time up to $n=7$ (I have not tested any higher values).  Equivalently, if you multiply both sides by $n$ then subtract $1$, you see that all the reciprocals sum to $0.5n-1$.  
I don't know where to start with generalizing this pattern as I have never seen explicit formulas for such sums, so I wanted to see if anyone knew if this was the case for all $n$ and how one could prove or disprove it.
 A: If you factor it as $\frac{1}{n}(1+\frac{1}{2})(1+\frac{1}{3})\dots (1+\frac{1}{n-1})$ it telescopes nicely.
A: With more systematic notation, your observation is that
$$ f(n) = \sum_{A\subseteq\{2,3,\ldots,n-1\}}\;\prod_{k\in A} \frac1k = \frac12n $$
(I don't see where you get "$0.5n -1$" out of it; subtracting $1$ doesn't seem to match anything in your examples.)
This does hold for all $n\ge 2$, and we can prove it by mathematical induction on $n$:
When we go from $f(n)$ to $f(n+1)$, the difference is that we now have more $A$s to sum over -- namely, we have all the ones we have before, plus all of the subsets that contain $n$. But each of the new subsets arises as one of the old subsets with $n$ appended, so we can write it as
$$ \begin{align} f(n+1) &= \sum_{A\subseteq\{2,3,\ldots,n-1\}}\; \prod_{k\in A} \frac1k + \sum_{A\subseteq\{2,3,\ldots,n-1\}}\;\frac1n \prod_{k\in A} \frac1k \\
&= f(n) + \frac1n f(n)
\\& = \frac{n+1}{n} f(n)
\\& = \frac{n+1}{n} \frac12 n
\\& = \dfrac12 (n+1) \end{align} $$
which is what is needed for the induction step.
A: Let $s_n$ be the sum of $1$ and the fractions for the $n$ case, e.g.
$$s_4 = 1+\frac{1}{2} + \frac{1}{3} + \frac{1}{2\cdot3}$$
Assume $s_k = k/2$ is true for some $k$.
For the $n=k+1$ case, 
$$s_{k+1} = s_k\cdot \left(1+\frac1k\right) = s_n\cdot\frac{k+1}{k} = \frac{k+1}{2}$$
For the $n=2$ case,
$$s_2 = 1 = \frac{2}{2}$$
By induction, $s_n = n/2$ is true for natural numbers $n\ge 2$. i.e.
$$\frac{1}{n}s_n = \frac12$$

Some example of the recursion $s_{k+1} = s_k\cdot\left(1+\frac1k\right) $:
$$\begin{align*}
s_4 &= 1+\frac{1}{2} + \frac{1}{3} + \frac{1}{2\cdot3}\\
&= \left(1 + \frac{1}{2}\right) + \frac{1}{3}\left(1 + \frac{1}{2}\right)\\
s_5 &= 1 + \frac12+\frac13+\frac1{2\cdot3}+\frac14+\frac1{2\cdot4} +\frac1{3\cdot4} + \frac{1}{2\cdot3\cdot4}\\
&= \left(1 + \frac12+\frac13+\frac1{2\cdot3}\right) + \frac14\left(1 + \frac12+\frac13+\frac1{2\cdot3}\right)
\end{align*}$$
