Riemann Zeta Function of 6
Theorem
The Riemann zeta function of $6$ is given by:
\(\ds \map \zeta 6\) | \(=\) | \(\ds \dfrac 1 {1^6} + \dfrac 1 {2^6} + \dfrac 1 {3^6} + \dfrac 1 {4^6} + \cdots\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\pi^6} {945}\) | ||||||||||||
\(\ds \) | \(\approx\) | \(\ds 1 \cdotp 01734 \, 3 \ldots\) |
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.
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.21$
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Bernoulli number
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {A}.21$: Euler ($\text {1707}$ – $\text {1783}$)
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.7$: Harmonic Numbers: $(7)$
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): zeta function