mathschallenge.net logo

Triangle Reciprocals

Problem

Find the value of the series of reciprocals of triangle numbers:

1/1 + 1/3 + 1/6 + 1/10 + 1/15 + ...


Solution

Let S = 1/1 + 1/3 + 1/6 + 1/10 + ...

therefore S/2 = 1/2 + 1/6 + 1/12 + 1/20 + ...
  = 1/(1times2) + 1/(2times3) + 1/(3times4) + ... + 1/(n(n + 1)) + ...
But 1/(n(n + 1)) = (n + 1 minus n)/(n(n + 1))
  = (n + 1)/(n(n + 1)) minus n/(n(n + 1))
  = 1/n minus 1/(n + 1)

Therefore S/2 = (1/1 minus 1/2) + (1/2 minus 1/3) + (1/3 minus 1/4) + ... = 1.

Hence S = 1/1 + 1/3 + 1/6 + ... = 2.

Problem ID: 236 (02 Aug 2005)     Difficulty: 3 Star

Only Show Problem