1. As n approaches infinity, so do n+1, n+2, ..., 2n, thus 1/(n+1), 1/(n+2), ..., 1/2n all approach zero; therefore the sum Xn approaches zero.
2. Each term 1/k(k+1)(k+2) can be written as 1/k * (1/(k+1) - 1/(k+2), so the sum equals:
ee[/colour]1/1 * (1/2 - 1/3) + 1/2 * (1/3 - 1/4) + ...
=...