Riemann Zeta Function of 6/Proof 2

Proof
Equating the product with the sum:

Each squared term in the product selected once:

Each unique combination of two squared terms in the product selected:


 * $\dfrac {x^4} {5!} = \dfrac {x^4} { \pi^4}

\paren {\paren 1 \paren {\dfrac 1 4} + \paren 1 \paren {\dfrac 1 9} + \paren 1 \paren {\dfrac 1 {16}} + \cdots + \paren {\dfrac 1 4} \paren {\dfrac 1 9} + \paren {\dfrac 1 4} \paren {\dfrac 1 {16}} + \cdots + \paren {\dfrac 1 9} \paren {\dfrac 1 {16} } }$ Term raised to the fourth power used to calculate Riemann Zeta Function of $4$

Each unique combination of three squared terms in the product selected:
 * $-\dfrac {x^6} {7!} = -\dfrac {x^6} {\pi^6}

\paren {\paren 1 \paren {\dfrac 1 4} \paren {\dfrac 1 9} + \paren 1 \paren {\dfrac 1 4} \paren {\dfrac 1 {16} } + \cdots + \paren 1 \paren {\dfrac 1 9} \paren {\dfrac 1 {16} } + \cdots + \paren {\dfrac 1 4} \paren {\dfrac 1 9} \paren {\dfrac 1 {16} } + \cdots}$ Term raised to the sixth used power to calculate Riemann Zeta Function of $6$

When we take the cube of a sum, we have:

Let $A = \dfrac 1 {1^2}, B = \dfrac 1 {2^2}, C = \dfrac 1 {3^2}, \cdots $

Then the becomes:
 * $\paren {\paren {\dfrac 1 {1^2} } + \paren {\dfrac 1 {2^2} } + \paren {\dfrac 1 {3^2} } + \cdots}^3 = \paren {\map \zeta 2}^3$

and the first term on the becomes:
 * $\paren {\paren {\dfrac 1 {1^2} }^3 + \paren {\dfrac 1 {2^2} }^3 + \paren {\dfrac 1 {3^2} }^3 + \cdots} = \map \zeta 6$

To make sense of the remaining two terms on the, we need:


 * $\paren {AB + AC + BC + \cdots} = \dfrac {\pi^4} {5!} $


 * $\paren {A + B + C + \cdots} = \dfrac {\pi^2} {3!} $

Finally, we have: