# 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. $\left(3 , 4\right)$=>1/(3*4)^2=>1/144) Given that the order of the numbers do not matter ($\left(3 , 4\right)$ and $\left(4 , 3\right)$ are the same), what is the sum of all Stephen fractions if there is a sum? Note that $\frac{1}{3 \cdot 4} ^ 2$ and $\frac{1}{2 \cdot 6} ^ 2$ 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!

Aug 8, 2018

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
$S = {\sum}_{x = 2}^{\infty} {\sum}_{y = 2}^{\infty} \frac{1}{x \cdot y} ^ 2$

$S = {\sum}_{x = 2}^{\infty} {\sum}_{y = 2}^{\infty} \frac{1}{{x}^{2} {y}^{2}}$

I replace in each sum $k = 2$ by $k = 1$, substracting to this sum the first missing element, knowing that ${\sum}_{z = 1}^{\infty} \frac{1}{{z}^{2}} = {\pi}^{2} / 6$,

$S = {\sum}_{x = 2}^{\infty} \frac{1}{x} ^ 2 {\sum}_{y = 2}^{\infty} \frac{1}{y} ^ 2$

$= {\sum}_{x = 1}^{\infty} \left(\frac{1}{x} ^ 2 - 1\right) {\sum}_{y = 2}^{\infty} \frac{1}{y} ^ 2$

$= {\sum}_{x = 1}^{\infty} \frac{1}{x} ^ 2 {\sum}_{y = 2}^{\infty} \frac{1}{y} ^ 2 - {\sum}_{y = 2}^{\infty} \frac{1}{y} ^ 2$

$= {\pi}^{2} / 6 \left({\sum}_{y = 1}^{\infty} \frac{1}{y} ^ 2 - 1\right) - \left({\sum}_{y = 1}^{\infty} \frac{1}{y} ^ 2 - 1\right)$

$= {\pi}^{2} / 6 \left({\pi}^{2} / 6 - 1\right) - \left({\pi}^{2} / 6 - 1\right)$

$= {\pi}^{4} / 36 - {\pi}^{2} / 6 - {\pi}^{2} / 6 + 1$

$S = 1 + {\pi}^{4} / 36 - {\pi}^{2} / 3$