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

The rth triangular is r(r+1)/2 and it's reciprocal is 2/[r(r+1)]=2*[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} + \dots + \frac{2}{n (n + 1)}

= 2 {\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \dots + \frac{1}{n (n + 1)}}

= 2 {[\frac{1}{1} - \frac{1}{2}] + [\frac{1}{2} - \frac{1}{3}] + \dots + [\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.