Sum of reciprocals of the triangle numbers 
Consider the sum of $n$ terms :
$S_n = 1 + \frac{1}{1+2} + \frac {1}{1+2+3} + ... + \frac {1}{1+2+3+...+n}$
  for $n \in N$.
Find the least rational number $r$ such that $S_n < r$, for all $n \in N$.

My attempt :
$S_n = 2(1-\frac{1}{2} + \frac {1}{2} - \frac{1}{3} + .... + \frac {1}{n} - \frac {1}{n+1}) = 2(1 - \frac {1}{n+1}) $
Now what to do with that '$r$' thing ?
How to proceed ?

Let $r=2$, and we can see that
$$\frac{2n-2}n=2-\frac2n<2$$
Similarly, as $n\to\infty$, the limit is $2$, so this is the least rational number satisfying the inequality.
