Riemann Zeta Function of 4/Proof 5

Theorem

The Riemann zeta function of $4$ is given by:

 $\ds \map \zeta 4$ $=$ $\ds \dfrac 1 {1^4} + \dfrac 1 {2^4} + \dfrac 1 {3^4} + \dfrac 1 {4^4} + \cdots$ $\ds$ $=$ $\ds \dfrac {\pi^4} {90}$ $\ds$ $\approx$ $\ds 1 \cdotp 08232 \, 3 \ldots$

Proof

From the Basel Problem we have:

$\displaystyle \map \zeta 2 = \sum_{n \mathop = 1}^{\infty} {\frac 1 {n^2}} = \frac {\pi^2} 6$

Therefore:

 $\ds \map \zeta 2$ $=$ $\ds \paren {\frac 1 {1^2 } + \frac 1 {2^2 } + \frac 1 {3^2 } + \frac 1 {4^2 } + \cdots }$ $\ds$ $=$ $\ds \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots }$

Taking the square of $\map \zeta 2$, we have:

 $\ds \displaystyle \paren {\map \zeta 2}^2$ $=$ $\ds \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots } \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots }$ $\ds$ $=$ $\ds \paren {\sum_{i \mathop = 1}^{\infty} {\frac 1 {i^2 } } } \paren {\sum_{j \mathop = 1}^{\infty} {\frac 1 {j^2 } } }$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^{\infty} \sum_{j \mathop = 1}^{\infty} {\frac 1 {i^2 } } {\frac 1 {j^2 } }$ Product of Absolutely Convergent Series $\ds$ $=$ $\ds \sum_{i \mathop = 1}^{\infty} {\frac 1 {i^4 } } + \sum_{i \mathop = 2}^{\infty} \sum_{j \mathop = 1}^{i-1} {\frac 1 {i^2 } } {\frac 1 {j^2 } } + \sum_{j \mathop = 2}^{\infty} \sum_{i \mathop = 1}^{j - 1} {\frac 1 {i^2 } } {\frac 1 {j^2 } }$ $\paren {i = j } + \paren {j \lt i } + \paren {j \gt i }$ $\ds$ $=$ $\ds \map \zeta 4 + 2 \dfrac {\pi^4} {5!}$ The sums $\paren {j < i }$ and $\paren {j > i }$ are symmetric and each equal to the coefficient of the 5th power term in the sin(x) expansion See Basel_Problem/Proof_2

$\begin{array}{r|cccccccccc} \displaystyle \paren {\map \zeta 2}^2 & \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {16}} & \cdots \\ \hline \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {1}}^2 & \paren {\dfrac {1} {1}} \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {1}} \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {1}} \paren {\dfrac {1} {16}} & \cdots \\ \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {4}} \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {4}}^2 & \paren {\dfrac {1} {4}} \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {4}} \paren {\dfrac {1} {16}} & \cdots \\ \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {9}} \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {9}} \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {9}}^2 & \paren {\dfrac {1} {9}} \paren {\dfrac {1} {16}} & \cdots \\ \paren {\dfrac {1} {16}} & \paren {\dfrac {1} {16}} \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {16}} \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {16}} \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {16}}^2 & \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ \end{array}$

 $\ds \map \zeta 4$ $=$ $\ds \paren {\map \zeta 2}^2 - 2 \dfrac {\pi^4} {5!}$ Rearranging $\ds$ $=$ $\ds \dfrac {\pi^4} {36} - \dfrac {\pi^4} {60}$ simplifying $\ds$ $=$ $\ds \dfrac {\pi^4} {90}$ simplifying

$\blacksquare$