40-00-TrickToSumAllTheReciprocalTriangularNumbers.tex

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\pmtitle{trick to sum all the reciprocal triangular numbers}
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\begin{document}
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 
$$1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + 
  \frac{1}{28} + \frac{1}{36} + \frac{1}{45} + \frac{1}{55} + \frac{1}{66} + 
  \frac{1}{78} +...=2$$



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