# Riemann Zeta Function of 6

## Theorem

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

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

## 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