Riemann Zeta Function of 6/Proof 2

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Theorem

The Riemann zeta function of $6$ is given by:

\(\displaystyle \map \zeta 6\) \(=\) \(\displaystyle \dfrac 1 {1^6} + \dfrac 1 {2^6} + \dfrac 1 {3^6} + \dfrac 1 {4^6} + \cdots\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\pi^6} {945}\)
\(\displaystyle \) \(\approx\) \(\displaystyle 1 \cdotp 01734 \, 3 \ldots\)


Proof

\(\displaystyle \sin x\) \(=\) \(\displaystyle x \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }\) Euler Formula for Sine Function
\(\displaystyle \displaystyle \dfrac {\sin x} x\) \(=\) \(\displaystyle \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} } = \paren {1 - \dfrac {x^2} {1^2 \pi^2} } \paren {1 - \dfrac {x^2} {2^2 \pi^2} } \paren {1 - \dfrac {x^2} {3^2 \pi^2} } \cdots\)


\(\displaystyle \sin x\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \dfrac {x^3} {3!} + \dfrac {x^5} {5!} - \dfrac {x^7} {7!} + \cdots\) Power Series Expansion for Sine Function
\(\displaystyle \displaystyle \dfrac {\sin x} x\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {x^{2 n} } {\paren {2 n + 1}!} = 1 - \dfrac {x^2} {3!} + \dfrac {x^4} {5!} - \dfrac {x^6} {7!} + \cdots\)


Equating the product with the sum:

\(\displaystyle \displaystyle \dfrac {\sin x} x\) \(=\) \(\displaystyle \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }\) \(\displaystyle = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n + 1}!}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {1 - \dfrac {x^2} {1 \pi^2} } x \paren {1 - \dfrac {x^2} {4 \pi^2} } x \paren {1 - \dfrac {x^2} {9 \pi^2} } x \cdots\) \(\displaystyle = 1 - \dfrac {x^2} {3!} + \dfrac {x^4} {5!} - \dfrac {x^6} {7!} + \cdots\)


Each squared term in the product selected once:

\(\displaystyle \displaystyle -\dfrac {x^2} {3!}\) \(=\) \(\displaystyle -\dfrac {x^2} {\pi^2} \paren {1 + \dfrac 1 4 + \dfrac 1 9 + \dfrac 1 {16} + \cdots}\) Term raised to the second power used to calculate the Riemann Zeta Function of $2$


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 $6$


\(\displaystyle \map \zeta 2\) \(=\) \(\displaystyle \dfrac 1 {1^2} + \dfrac 1 {2^2} + \dfrac 1 {3^2} + \dfrac 1 {4^2} + \cdots\) \(\displaystyle = \dfrac {\pi^2} 6\) Basel Problem
\(\displaystyle \paren {\map \zeta 2}^3\) \(=\) \(\displaystyle \paren {\dfrac 1 {1^2} + \dfrac 1 {2^2} + \dfrac 1 {3^2} + \dfrac 1 {4^2} + \cdots}^3\) \(\displaystyle = \paren {\dfrac {\pi^2} 6}^3 = \dfrac {\pi^6} {216}\)

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

\(\displaystyle \paren {A + B + C + \cdots}^3\) \(=\) \(\displaystyle \paren {A^3 + B^3 + C^3 + \cdots} + 3 \paren {A^2 B + A B^2 + A^2 C + A C^2 + B^2 C + B C^2 + \cdots} + 6 \paren {A B C + \cdots}\)

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


Then the left hand side 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 right hand side 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 right hand side, we need:

$\paren {AB + AC + BC + \cdots} = \dfrac {\pi^4} {5!} $
$\paren {A + B + C + \cdots} = \dfrac {\pi^2} {3!} $
\(\displaystyle \paren {AB + AC + BC + \cdots} \paren {A + B + C + \cdots}\) \(=\) \(\displaystyle \paren {A^2B + AB^2 + A^2C + AC^2 + B^2C + BC^2 + \cdots} + 3\paren {ABC + \cdots}\)
\(\displaystyle 3 \paren {AB + AC + BC + \cdots} \paren {A + B + C + \cdots}\) \(=\) \(\displaystyle 3 \paren {A^2B + AB^2 + A^2C + AC^2 + B^2C + BC^2 + \cdots} + 9 \paren {ABC + \cdots}\) We have 3 too many of the 'ABC' type (only need 6) - need to subtract 3 of these

Finally, we have:

\(\displaystyle \paren {\map \zeta 2}^3\) \(=\) \(\displaystyle \map \zeta 6 + 3 \dfrac {\pi^2} {3!} \dfrac {\pi^4} {5!} - 3 \dfrac {\pi^6} {7!}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map \zeta 6\) \(=\) \(\displaystyle \paren {\map \zeta 2}^3 - 3 \dfrac {\pi^2} {3!} \dfrac {\pi^4} {5!} + 3 \dfrac {\pi^6} {7!}\) rearranging
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\pi^6} {216} - 21 \dfrac {\pi^6} {7!} + 3 \dfrac {\pi^6} {7!}\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle 70 \dfrac {\pi^6} {3 x 7!} - 54 \dfrac {\pi^6} {3 x 7!}\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle 16 \dfrac {\pi^6} {3 x 7!}\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\pi^6} {945}\)

$\blacksquare$