1/(1+2+...+n) = 2/(n)(n+1) = 2[1/n - 1/(n+1)]
1+1/3+1/6+1/10+1/15+.+1/1+2+3+...+n
=2 (1/1-1/2+1/2-1/3+1/3-1/4 + ...+1/n-1/n+1)
=2 [1-1/(n+1)]
=2n/(n+1)