Inspired by a BMO Question

The formula for the triangle numbers is $T_n = \frac{n(n + 1)}{2}$ and so $\sum_{n=1}^{\infty} \frac{1}{T_n} = 2\sum_{n=1}^{\infty} \frac{1}{n(n + 1)}$ . Now using partial fractions: $x(n + 1) + yn = 1$ Plugging in $n = 0$ : $x = 1$ Plugging in $n = -1$ : $-y = 1, y = -1$ Indeed, $n + 1 - n = 1$ . Going back to our original summation, $2\sum_{n = 1}^{\infty} \frac{1}{n(n + 1)} = 2(\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} ... )$ From this telescoping series it is clear that $2\sum_{n = 1}^{k} \frac{1}{n(n + 1)} = 2 - 2\frac{1}{n + 1}$ . It is obvious that as $n$ tends to $\infty$ , so does $n + 1$ , and as $n + 1$ tends to $\infty \frac{1}{n + 1}$ tends to $0$ . Hence $2\sum_{n=1}^{\infty} \frac{1}{n(n + 1)}$ = 2