Riemann Zeta Function of 6/Proof 2

Proof
From Euler Formula for Sine Function:


 * $\displaystyle \sin x = x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }$

Dividing by $x$:


 * $(1): \quad \displaystyle \dfrac {\sin x} x = \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} } = \paren {1 - \frac {x^2} {1^2 \pi^2} } \paren {1 - \frac {x^2} {2^2 \pi^2} } \paren {1 - \frac {x^2} {3^2 \pi^2} } \cdots$

From Power Series Expansion for Sine Function:


 * $\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7} {7!} + \cdots$

Dividing by $x$:


 * $(2): \quad \displaystyle \dfrac {\sin x} x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n + 1}!} = 1 - \frac {x^2} {3!} + \frac {x^4} {5!} - \frac {x^6} {7!} + \cdots$

Equating $(1)$ with $(2)$, we have:


 * $\displaystyle \dfrac {\sin x} x = \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} } = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n + 1}!}$


 * $\dfrac {\sin x} x = \paren {1 - \dfrac {x^2} {1 \pi^2} } x \paren {1 - \dfrac {x^2} {4 \pi^2} } x \paren {1 - \dfrac {x^2} {9 \pi^2} } x \cdots = 1 - \dfrac {x^2} {3!} + \dfrac {x^4} {5!} - \dfrac {x^6} {7!} + \cdots$

Each squared term in $(1)$ selected once:
 * $-\dfrac {x^2} {3!} = -\dfrac {x^2} {\pi^2} \paren {1 + \dfrac 1 4 + \dfrac 1 9 + \dfrac 1 {16} + \cdots} $"Choose 1" - used to calculate the Riemann Zeta Function of $2$

Every possible combination of two squared terms in $(1)$ selected once:
 * $\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} } }$ "Choose 2" - used to calculate Riemann Zeta Function of $4$

Every possible combination of three squared terms in the infinite product selected once:
 * $-\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}$ "Choose 3" - required here to calculate Riemann Zeta Function of $6$

From the Basel Problem:

Therefore:

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^6} {5!} $


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

Finally, we have: