This solution was sent by Etienne Chan, age 15, 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!

1 r −  1 r+1 =  (r+1)−r r(r+1) =  1 r(r+1)

The sum of the reciprocals of the first n triangular numbers

2 1 ×2 +  2 2 ×3 + …+  2 n(n+1)

= 2 {  1 1 ×2 +  1 2 ×3 + …+  1 n(n+1) }

= 2 { [  1 1 −  1 2 ] + [  1 2 −  1 3 ] + …+ [  1 n −  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 −  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.