Sum of the Reciprocals of Triangular Numbers

This problem involves finding the sum of the reciprocals of a series of consecutive triangular numbers. The solution relies on knowing the starting term number ($n$) and the ending term number ($m$) of the triangular numbers in the sequence.

$\frac{1}{T_n} + \frac{1}{T_{n+1}} + \dots + \frac{1}{T_m} = 2(\frac{1}{n} - \frac{1}{m+1})$

For example, $\frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21}$ corresponds to the sum from the 3rd to the 6th triangular number. So, $n=3$ and $m=6$. The sum is $2(\frac{1}{3} - \frac{1}{7}) = \frac{8}{21}$.