Is the sum of reciprocals of all products from $2$ to $n-1$ always $0.5n-1$?
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.
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.