An interesting problem...?

Stephen is making "the Stephen fractions". He chooses a set of two whole numbers (could be identical) greater than 1. He multiplies them, squares the result, and finds the reciprocal of the answer. (e.g. #(3,4)#=>#1/(3*4)^2=>1/144)# Given that the order of the numbers do not matter (#(3,4)# and #(4,3)# are the same), what is the sum of all Stephen fractions if there is a sum? Note that #1/(3*4)^2# and #1/(2*6)^2# are counted as two different fractions.

I think I have found a solution, and I would like to know how you would solve this problem!

1 Answer
Aug 8, 2018

Answer:

The sum of all Stephen fractions is #S=1+pi^4/36-pi^2/3#

Explanation:

Evaluating the problem, we know that Stephen fractions sum
#S= sum_(x=2)^(oo)sum_(y=2)^(oo)1/(x*y)^2#

#S=sum_(x=2)^(oo)sum_(y=2)^(oo)1/(x^2y^2)#

I replace in each sum #k=2# by #k=1#, substracting to this sum the first missing element, knowing that #sum_(z=1)^(oo)1/(z^2)=pi^2/6#,

#S=sum_(x=2)^(oo)1/x^2sum_(y=2)^(oo)1/y^2#

#=sum_(x=1)^(oo)(1/x^2-1)sum_(y=2)^(oo)1/y^2#

#=sum_(x=1)^(oo)1/x^2sum_(y=2)^(oo)1/y^2-sum_(y=2)^(oo)1/y^2#

#=pi^2/6(sum_(y=1)^(oo)1/y^2-1)-(sum_(y=1)^(oo)1/y^2-1)#

#=pi^2/6(pi^2/6-1)-(pi^2/6-1)#

#=pi^4/36-pi^2/6-pi^2/6+1#

#S=1+pi^4/36-pi^2/3#

\0/ Here's our answer !