Home > Quick Problems > Algebra

Note: this page uses MathJax for formatting, so may take a few seconds to load.

function showSolution() {
if (document.getElementById('divSolution').style.display == "none")
{ document.getElementById('divSolution').style.display = "block"; }
else
{ document.getElementById('divSolution').style.display = "none"; }
}
document.getElementById('divSolution').style.display = "none";

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 $2$.