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

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

Title	trick to sum all the reciprocal triangular numbers
Canonical name	TrickToSumAllTheReciprocalTriangularNumbers
Date of creation	2013-03-22 18:58:21
Last modified on	2013-03-22 18:58:21
Owner	juanman (12619)
Last modified by	juanman (12619)
Numerical id	6
Author	juanman (12619)
Entry type	Result
Classification	msc 40-00
Classification	msc 11A99
Related topic	TriangularNumbers