mathschallenge.net

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 + ...

S/2	=	1/2 + 1/6 + 1/12 + 1/20 + ...
	=	1/(12) + 1/(23) + 1/(34) + ... + 1/(n(n + 1)) + ...

But 1/(n(n + 1))	=	(n + 1 n)/(n(n + 1))
	=	(n + 1)/(n(n + 1)) n/(n(n + 1))
	=	1/n 1/(n + 1)

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

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