ProofWiki:Sandbox

Proof
From the Basel Problem/Proof 11, we have:


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


 * $\displaystyle $


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


 * $\displaystyle $

Each squared term in the infinite product selected once:


 * $\displaystyle $

Every possible combination of two squared terms in the infinite product selected once:


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

\paren {\paren {1} \paren {\frac 1 4} + \paren {1} \paren {\frac 1 9} + \paren {1} \paren {\frac 1 {16}} + \cdots + \paren {\frac 1 4} \paren {\frac 1 9} + \paren {\frac 1 4} \paren {\frac 1 {16}} + \cdots + \paren {\frac 1 9} \paren {\frac 1 {16}}} $"Choose 2" - used to calculate Riemann Zeta Function of $4$


 * $\displaystyle $

Every possible combination of three squared terms in the infinite product selected once:


 * $\displaystyle - \frac {x^6} {7!} = - \frac {x^6} { \pi^6}

\paren {\paren {1} \paren {\frac {1} {4}} \paren {\frac {1} {9}} + \paren {1} \paren {\frac {1} {4}} \paren {\frac {1} {16}} + \cdots + \paren {1} \paren {\frac {1} {9}} \paren {\frac {1} {16}} + \cdots + \paren {\frac {1} {4}} \paren {\frac 1 9} \paren {\frac 1 {16}} + \cdots} $"Choose 3" - required here to calculate Riemann Zeta Function of $6$


 * $\displaystyle $

From the Basel Problem, we have:

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 LHS becomes:
 * $\displaystyle \paren {\paren{\frac 1 {1^2}} + \paren{\frac 1 {2^2}} + \paren{\frac 1 {3^2}} + \cdots}^3 = \paren {\map \zeta 2}^3$

And the first term on the RHS becomes:
 * $\displaystyle \paren {\paren{\frac 1 {1^2}}^3 + \paren{\frac 1 {2^2}}^3 + \paren{\frac 1 {3^2}}^3 + \cdots}^3 = \map \zeta 6$