trick to sum all the reciprocal triangular numbers

The following trick to sum all the reciprocals of the triangular numbers is funny:

$\sigma$$=$$1+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{6}}+{\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{15}}+{\displaystyle \frac{1}{21}}+{\displaystyle \frac{1}{28}}+{\displaystyle \frac{1}{36}}+{\displaystyle \frac{1}{45}}+{\displaystyle \frac{1}{55}}+{\displaystyle \frac{1}{66}}+{\displaystyle \frac{1}{78}}+\mathrm{\dots }$

$=$$1+\left({\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{6}}\right)+\left({\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{15}}\right)+\left({\displaystyle \frac{1}{21}}+{\displaystyle \frac{1}{28}}\right)+\left({\displaystyle \frac{1}{36}}+{\displaystyle \frac{1}{45}}\right)+\left({\displaystyle \frac{1}{55}}+{\displaystyle \frac{1}{66}}\right)+\mathrm{\dots }$

$=$$1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{6}}+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{15}}+\mathrm{\dots }$

$=$$1+{\displaystyle \frac{1}{2}}\left(1+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{6}}+{\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{15}}+\mathrm{\dots }\right)$

which implies $\sigma =1+\sigma /2$ and hence

$$1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}+\frac{1}{78}+\mathrm{\dots }=2$$