Riemann Zeta Function of 4/Proof 5

Proof
Create a multiplication table where the column down the and the row across the top each contains the terms of zeta function of $2$:


 * $\begin {array} {c|cccccccccc}

\paren {\map \zeta 2}^2 & \paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {2^2} } & \paren {\dfrac 1 {3^2} } & \paren {\dfrac 1 {4^2} } & \cdots \\ \hline

\paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {1^4} } & \paren {\dfrac 1 {1^2} } \paren {\dfrac 1 {2^2} } & \paren {\dfrac 1 {1^2} } \paren {\dfrac 1 {3^2} } & \paren {\dfrac 1 {1^2} } \paren {\dfrac 1 {4^2} } & \cdots \\

\paren {\dfrac 1 {2^2} } & \paren {\dfrac 1 {2^2} } \paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {2^4} } & \paren {\dfrac 1 {2^2} } \paren {\dfrac 1 {3^2} } & \paren {\dfrac 1 {2^2} } \paren {\dfrac 1 {4^2} } & \cdots \\

\paren {\dfrac 1 {3^2} } & \paren {\dfrac 1 {3^2} } \paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {3^2} } \paren {\dfrac 1 {2^2} } & \paren {\dfrac 1 {3^4} } & \paren {\dfrac 1 {3^2} } \paren {\dfrac 1 {4^2} } & \cdots \\

\paren {\dfrac 1 {4^2} } & \paren {\dfrac 1 {4^2} } \paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {4^2} } \paren {\dfrac 1 {2^2} } & \paren {\dfrac 1 {4^2} } \paren {\dfrac 1 {3^2} } & \paren {\dfrac 1 {4^4} } & \cdots \\

\vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end {array}$

The sum of all of the entries in this table is equal to $\paren {\map \zeta 2}^2$.


 * $\map \zeta 4$ is the sum of the entries along the main diagonal.

We have:

From Proof $2$, the coefficient of the $x^5$ term is $\ds \dfrac 1 {\pi^4} \sum_{n_1 \mathop = 1}^{k - 1} \sum_{n_2 \mathop = {n_1 + 1} }^k \paren {\frac 1 {n_1^2} } \paren {\frac 1 {n_2^2} }$

Equating the coefficient of $x^5$ in the Power Series Expansion for Sine Function with the Euler Formula for Sine Function, we have:


 * $\ds \lim_{k \mathop \to \infty} \dfrac 1 {\pi^4} \sum_{n_1 \mathop = 1}^{k - 1} \sum_{n_2 \mathop = {n_1 + 1} }^k \paren {\frac 1 {n_1^2} } \paren {\frac 1 {n_2^2} } = \frac 1 {5!}$

Therefore:


 * $\ds \sum_{n_1 \mathop = 1}^\infty \sum_{n_2 \mathop = {n_1 + 1} }^\infty \paren {\frac 1 {n_1^2} } \paren {\frac 1 {n_2^2} } = \frac {\pi^4} {5!}$

Therefore: