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!


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
 + ¼+ 2
n(n+1)

= 2 { 1
1 ×2
 + 1
2 ×3
 + ¼+ 1
n(n+1)
 }

= 2 { [ 1
1
 - 1
2
 ] + [ 1
2
 - 1
3
 ] + ¼+ [ 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.