Riemann Zeta Function of 6

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Theorem

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

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

This sequence is A013664 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof 1

By Fourier Series: $x^6$ over $-\pi$ to $\pi$, for $x \in \closedint {-\pi} \pi$:

$\ds x^6 = \frac {\pi^6} 7 + \sum_{n \mathop = 1}^\infty \frac {12 n^4 \pi^4 - 240 n^2 \pi^2 +1440} {n^6} \, \map \cos {n \pi} \, \map \cos {n x}$

Setting $x = \pi$:

\(\ds \pi^6\) \(=\) \(\ds \frac {\pi^6} 7 + \sum_{n \mathop = 1}^\infty \frac {12 n^4 \pi^4 - 240 n^2 \pi^2 +1440} {n^6} \, \map {\cos^2} {n \pi}\)
\(\ds \leadsto \ \ \) \(\ds \frac { 6 \pi^6} 7\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \frac {12 \pi^4} {n^2} - \sum_{n \mathop = 1}^\infty \frac {240 \pi^2} {n^4} + \sum_{n \mathop = 1}^\infty \frac {1440} {n^6}\) Cosine of Multiple of Pi
\(\ds \leadsto \ \ \) \(\ds \frac {\pi^6} 7\) \(=\) \(\ds 2 \pi^4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} - 40 \pi^2 \sum_{n \mathop = 1}^\infty \frac 1 {n^4} + 240 \sum_{n \mathop = 1}^\infty \frac 1 {n^6}\)
\(\ds \) \(=\) \(\ds - \frac {\pi^6} 9 + 240 \sum_{n \mathop = 1}^\infty \frac 1 {n^6}\) Basel Problem and Riemann Zeta Function of 4
\(\ds \leadsto \ \ \) \(\ds 240 \sum_{n \mathop = 1}^\infty \frac 1 {n^6}\) \(=\) \(\ds \frac {\pi^6} 9 + \frac {\pi^6} 7\) rearranging
\(\ds \) \(=\) \(\ds \frac {16 \pi^4} {63}\)
\(\ds \leadsto \ \ \) \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^6}\) \(=\) \(\ds \frac {\pi^6} {945}\)

$\blacksquare$


Proof 2

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


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


Dividing out the x factor on both sides and equating the product with the sum, we have:

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


Equating the $x^2$ term on both sides of the equation, we obtain the value of the sum of the individual terms in the Basel Problem:

\(\ds -\dfrac {x^2} {3!}\) \(=\) \(\ds -\dfrac {x^2} {\pi^2} \paren {1 + \dfrac 1 4 + \dfrac 1 9 + \dfrac 1 {16} + \cdots}\) Basel Problem
\(\ds \dfrac {\pi^2} 6\) \(=\) \(\ds \paren {1 + \dfrac 1 4 + \dfrac 1 9 + \dfrac 1 {16} + \cdots}\)


Equating the $x^4$ term on both sides of the equation, we obtain the value of the sum of the product of every unique combination of $2$ terms from the Basel Problem:

\(\ds \dfrac {x^4} {5!}\) \(=\) \(\ds \dfrac {x^4} {\pi^4} \paren {\paren 1 \paren {\dfrac 1 4 } + \paren 1 \paren {\dfrac 1 9 } + \paren 1 \paren {\dfrac 1 {16} } + \cdots + \paren {\dfrac 1 4} \paren {\dfrac 1 9} + \paren {\dfrac 1 4} \paren {\dfrac 1 {16} } + \cdots + \paren {\dfrac 1 9} \paren {\dfrac 1 {16} } + \cdots }\)
\(\ds \dfrac {\pi^4} {120 }\) \(=\) \(\ds \paren {\paren 1 \paren {\dfrac 1 4 } + \paren 1 \paren {\dfrac 1 9 } + \paren 1 \paren {\dfrac 1 {16} } + \cdots + \paren {\dfrac 1 4} \paren {\dfrac 1 9} + \paren {\dfrac 1 4} \paren {\dfrac 1 {16} } + \cdots + \paren {\dfrac 1 9} \paren {\dfrac 1 {16} } + \cdots }\)


Equating the $x^6$ term on both sides of the equation, we obtain the value of the sum of the product of every unique combination of $3$ terms from the Basel Problem:

\(\ds -\dfrac {x^6} {7!}\) \(=\) \(\ds -\dfrac {x^6} {\pi^6} \paren {\paren 1 \paren {\dfrac 1 4} \paren {\dfrac 1 9} + \paren 1 \paren {\dfrac 1 4} \paren {\dfrac 1 {16} } + \cdots + \paren 1 \paren {\dfrac 1 9} \paren {\dfrac 1 {16} } + \cdots + \paren {\dfrac 1 4} \paren {\dfrac 1 9} \paren {\dfrac 1 {16} } + \cdots}\)
\(\ds \dfrac {\pi^6} {7!}\) \(=\) \(\ds \paren {\paren 1 \paren {\dfrac 1 4} \paren {\dfrac 1 9} + \paren 1 \paren {\dfrac 1 4} \paren {\dfrac 1 {16} } + \cdots + \paren 1 \paren {\dfrac 1 9} \paren {\dfrac 1 {16} } + \cdots + \paren {\dfrac 1 4} \paren {\dfrac 1 9} \paren {\dfrac 1 {16} } + \cdots}\)


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

\(\ds \paren {A + B + C + \cdots}^3\) \(=\) \(\ds \paren {A^3 + B^3 + C^3 + \cdots} + 3 \paren {A^2 B + A B^2 + A^2 C + A C^2 + B^2 C + B C^2 + \cdots} + 6 \paren {A B C + \cdots}\)
\(\ds \paren {\text {Cube of Sum } }\) \(=\) \(\ds \paren {\text {Sum of Cubes } } + 3 \paren { \text {Product of 3 Terms, with one term a perfect square (Every Combination) } } + 6 \paren { \text {Product of 3 Unique Terms (Every Combination) } }\)


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


Then the left hand side (Cube of Sum) becomes:

\(\ds \paren {\paren {\dfrac 1 {1^2} } + \paren {\dfrac 1 {2^2} } + \paren {\dfrac 1 {3^2} } + \cdots}^3\) \(=\) \(\ds \paren {\map \zeta 2}^3\)


and the first term on the right hand side (Sum of Cubes) becomes:

\(\ds \paren {\paren {\dfrac 1 {1^2} }^3 + \paren {\dfrac 1 {2^2} }^3 + \paren {\dfrac 1 {3^2} }^3 + \cdots}\) \(=\) \(\ds \map \zeta 6\)


To obtain the remaining two terms on the right hand side, we have:

\(\ds \paren {AB + AC + BC + \cdots}\) \(=\) \(\ds \dfrac {\pi^4} {5!}\) From the $x^4$ term above: Product of $2$ Unique terms
\(\ds \paren {A + B + C + \cdots}\) \(=\) \(\ds \dfrac {\pi^2} {3!}\) From the $x^2$ term above: Each term once
\(\ds \paren {AB + AC + BC + \cdots} \paren {A + B + C + \cdots}\) \(=\) \(\ds \paren {A^2B + AB^2 + A^2C + AC^2 + B^2C + BC^2 + \cdots} + 3\paren {ABC + \cdots}\)
\(\ds 3 \paren {AB + AC + BC + \cdots} \paren {A + B + C + \cdots}\) \(=\) \(\ds 3 \paren {A^2B + AB^2 + A^2C + AC^2 + B^2C + BC^2 + \cdots} + 9 \paren {ABC + \cdots}\) We have 3 too many of the 'ABC' type (only need 6) - need to subtract 3 of these


Finally, we have:

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

$\blacksquare$


Proof 3

\(\ds \sum_{n \mathop = 1}^{\infty} \frac 1 {n^6}\) \(=\) \(\ds \map \zeta 6\) Definition of Riemann Zeta Function
\(\ds \) \(=\) \(\ds \paren {-1}^4 \frac {B_6 2^5 \pi^6} {6!}\) Riemann Zeta Function at Even Integers
\(\ds \) \(=\) \(\ds \frac 1 {42} \cdot \frac {2^5 \pi^6} {6!}\) Definition of Sequence of Bernoulli Numbers
\(\ds \) \(=\) \(\ds \frac {32 \pi^6} {42 \cdot 720}\) Definition of Factorial
\(\ds \) \(=\) \(\ds \frac {\pi^6} {945}\)

$\blacksquare$


Historical Note

The Riemann Zeta Function of 6 was solved by Leonhard Euler, using the same technique as for the Riemann Zeta Function of 4 and the Basel Problem.


If only my brother were alive now.
-- Johann Bernoulli


Sources