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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. $(3, 4)$ $(3,4)$ => $\frac{1}{{(3 \cdot 4)}^{2}} \Rightarrow \frac{1}{144})$ $1/(34)^2=>1/144)$ Given that the order of the numbers do not matter ( $(3, 4)$ $(3,4)$ and $(4, 3)$ $(4,3)$ are the same), what is the sum of all Stephen fractions if there is a sum? Note that $\frac{1}{{(3 \cdot 4)}^{2}}$ $1/(34)^2$ and $\frac{1}{{(2 \cdot 6)}^{2}}$ $1/(2*6)^2$ 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!

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Answer 1 · 2018-08-08T10:34:08

The sum of all Stephen fractions is $S = 1 + \frac{π^{4}}{36} - \frac{π^{2}}{3}$ $S=1+pi^4/36-pi^2/3$

Explanation:

Evaluating the problem, we know that Stephen fractions sum
$S = \infty \sum x = 2 \infty \sum y = 2 \frac{1}{{(x \cdot y)}^{2}}$ $S= sum_(x=2)^(oo)sum_(y=2)^(oo)1/(x*y)^2$

$S = \infty \sum x = 2 \infty \sum y = 2 \frac{1}{x^{2} y^{2}}$ $S=sum_(x=2)^(oo)sum_(y=2)^(oo)1/(x^2y^2)$

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

$S = \infty \sum x = 2 \frac{1}{x^{2}} \infty \sum y = 2 \frac{1}{y^{2}}$ $S=sum_(x=2)^(oo)1/x^2sum_(y=2)^(oo)1/y^2$

$= \infty \sum x = 1 (\frac{1}{x^{2}} - 1) \infty \sum y = 2 \frac{1}{y^{2}}$ $=sum_(x=1)^(oo)(1/x^2-1)sum_(y=2)^(oo)1/y^2$

$= \infty \sum x = 1 \frac{1}{x^{2}} \infty \sum y = 2 \frac{1}{y^{2}} - \infty \sum y = 2 \frac{1}{y^{2}}$ $=sum_(x=1)^(oo)1/x^2sum_(y=2)^(oo)1/y^2-sum_(y=2)^(oo)1/y^2$

$= \frac{π^{2}}{6} (\infty \sum y = 1 \frac{1}{y^{2}} - 1) - (\infty \sum y = 1 \frac{1}{y^{2}} - 1)$ $=pi^2/6(sum_(y=1)^(oo)1/y^2-1)-(sum_(y=1)^(oo)1/y^2-1)$

$= \frac{π^{2}}{6} (\frac{π^{2}}{6} - 1) - (\frac{π^{2}}{6} - 1)$ $=pi^2/6(pi^2/6-1)-(pi^2/6-1)$

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

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

\0/ Here's our answer !

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