# trick to sum all the reciprocal triangular numbers

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

 $\displaystyle\sigma$ $\displaystyle=$ $\displaystyle 1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}% +\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}+\frac{1}{78}% +...$ $\displaystyle=$ $\displaystyle 1+\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{10}+\frac{% 1}{15}\right)+\left(\frac{1}{21}+\frac{1}{28}\right)+\left(\frac{1}{36}+\frac{% 1}{45}\right)+\left(\frac{1}{55}+\frac{1}{66}\right)+...$ $\displaystyle=$ $\displaystyle 1+\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{2}\cdot\frac{% 1}{6}+\frac{1}{2}\cdot\frac{1}{10}+\frac{1}{2}\cdot\frac{1}{15}+...$ $\displaystyle=$ $\displaystyle 1+\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{% 1}{15}+...\right)$

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

 $1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+% \frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}+\frac{1}{78}+...=2$
Title trick to sum all the reciprocal triangular numbers TrickToSumAllTheReciprocalTriangularNumbers 2013-03-22 18:58:21 2013-03-22 18:58:21 juanman (12619) juanman (12619) 6 juanman (12619) Result msc 40-00 msc 11A99 TriangularNumbers