Reciprocal triangles

This solution was sent by Etienne from Parramatta Highschool, NSW Australia.

The $r$th triangular is $r(r+1)/2$ and it's reciprocal is $2/[r(r+1)]=2 \times [1/(r(r+1))]$

Now $1/r(r+1) = 1/r - 1/(r+1)$. Take note of this, very useful technique!

$$\frac{1}{r}-\frac{1}{r+1}=\frac{(r+1)-r}{r(r+1)}=\frac{1}{r(r+1)}$$

The sum of the reciprocals of the first n triangular numbers

$$\frac{2}{1\times 2}+\frac{2}{2\times 3}+\cdots +\frac{2}{n(n+1)}$$ $$=2\{\frac{1}{1\times 2}+\frac{1}{2\times 3}+\cdots +\frac{1}{n(n+1)}\}$$ $$=2\{[\frac{1}{1}-\frac{1}{2}]+[\frac{1}{2}-\frac{1}{3}]+\cdots +[\frac{1}{n}-\frac{1}{(n+1)}]\}$$

Surprise, you get terms that cancel out each other, ie $-1/2$ and $1/2$, $-1/3$ and $1/3$, $-1/n$ and $1/n$. This is called 'telescoping'.

The sum thus equals $$2\{1-\frac{1}{(n+1)}\}$$

When n is large, $1/(n+1)$ is very small, so the sum is approximately $2$.

When $n$ tends to infinity, $1/(n+1)$ tends to $0$, and it turns out the infinite sum is exactly $2$.