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

\begin{eqnarray*} \sigma&=&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} +...\\ &=&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) +...\\ &=&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} +... \\ &=&1 + \frac{1}{2}\left( 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} +...\right) \end{eqnarray*}which implies $\sigma=1+\sigma/2$ and hence $$1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + \frac{1}{28} + \frac{1}{36} + \frac{1}{45} + \frac{1}{55} + \frac{1}{66} + \frac{1}{78} +...=2$$