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.

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.
