Triangles to Infinity

Relevant wiki: Telescoping Series - Sum

$\begin{aligned} S & = \sum_{r=1}^\infty \frac 1{\sum_{m=0}^r m} \\ & = \sum_{r=1}^\infty \frac 1{\frac {r(r+1)}2} \\ & = \sum_{r=1}^\infty \frac 2{r(r+1)} \\ & = 2 \sum_{r=1}^\infty \left(\frac 1r - \frac 1{r+1}\right) \\ & = 2 \left(\sum_{r=1}^\infty \frac 1r - \sum_{r=\color{#D61F06}{2}}^\infty \frac 1{\color{#D61F06}{r}} \right) \\ & = 2 \left(\frac 11 \right) = \boxed{2} \end{aligned}$

It is well known that the sum to r of triangle numbers is $\frac{r(r+1)}{2}$ one over this is $\frac{2}{r(r+1)}$ this can be rewritten as $2(\frac{1}{r}-\frac{1}{r+1})$ .

Each value of r will cancel with the last leaving the first an last terms as the only ones remaining, that is $n=2(\frac{1}{1}+\frac{1}{\infty})=2$