Mathacadabra
Home > Quick Problems > Algebra
Note: this page uses MathJax for formatting, so may take a few seconds to load.


A Sum of Reciprocals

The $n$th triangular number is given by the formula \[ 1+2+3+\cdots+n=\frac{n(n+1)}{2}, \quad \mbox{for} \; \, n \geq 1. \] Such a number is called triangular because it can be represented by a triangle with $1$ dot in the first row, $2$ dots in the second row, $3$ dots in the third row, $\ldots$, $n$ dots in the $n$th row. The sequence of triangular numbers looks like \[ 1, \; 3, \; 6, \; 10, \; 15, \; 21, \; \ldots . \]

What is the sum of the reciprocals of all the triangular numbers, \[ \frac{1}{1}+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\cdots \, ? \]

Solution
The trick is to write the reciprocal of a triangular number as a difference of fractions: \[ \frac{2}{n(n+1)}=\frac{2}{n}-\frac{2}{n+1}. \] Thus, the sum of the reciprocals of the triangular numbers is a telescoping series: \begin{align*} &\left(\frac{2}{1}-\frac{2}{2}\right)+\left(\frac{2}{2}-\frac{2}{3}\right)+\left(\frac{2}{3}-\frac{2}{4}\right)+\left(\frac{2}{4}-\frac{2}{5}\right)+\left(\frac{2}{5}-\frac{2}{6}\right)+\left(\frac{2}{6}-\frac{2}{7}\right)+\cdots \\ &=\frac{2}{1}+\left(-\frac{2}{2}+\frac{2}{2}\right)+\left(-\frac{2}{3}+\frac{2}{3}\right)+\left(-\frac{2}{4}+\frac{2}{4}\right)+\left(-\frac{2}{5}+\frac{2}{5}\right)+\left(-\frac{2}{6}+\frac{2}{6}\right)+\cdots\\ &= \frac{2}{1}+0+0+0+0+0+\cdots. \end{align*} The series collapses to $2$, so the sum of the reciprocals of the triangular numbers is $2$.

However, we must be careful with this type of argument. To be sure that the series converges, we must look at the partial sums: \[ \left(\frac{2}{1}-\frac{2}{2}\right)+\cdots+\left(\frac{2}{n}-\frac{2}{n+1}\right)=2-\frac{2}{n+1}. \] Since the $n$th partial sum converges to $2$, the series really sums to $2$.