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 ?

Solution 1:

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.

Solution 2:

First, let's point out that it isn't obvious why there is a "least" such rational number. For instance, if $S_n = 2$ for all $n$, then there isn't such a least rational number. So let's first prove that $r$ exists.

Let $A = \{x \in \Bbb Q: S_n < x, \forall n\}$. The question should read:

Prove that $\min A$ exists and find its value.

$$S_n = \sum_{k=1}^n \frac1{\sum_{t = 1}^k t} = \sum_{k=1}^n \frac1{\frac{k(k+1)}2} = 2\left( \sum_{k=1}^n \frac1k - \sum_{k=1}^n \frac1{k+1}\right) \\ = 2 \left(1 - \frac1{n+1} \right)$$

If $x \in A$, then $S_n < x, \forall n \implies \lim S_n \le x \implies 2 \le x$, so $2$ is a lower bound of $A$. On the other hand, $2 \in A$. Thus, $2 = \min A$; in particular, $\min A$ exists.