Theorem: Riemann Zeta Function of 4

The Riemann zeta function of $4$ is given by:
{{begin-eqn}}
{{eqn | l = \map \zeta 4
      | r = \dfrac 1 {1^4} +  \dfrac 1 {2^4} +  \dfrac 1 {3^4} +  \dfrac 1 {4^4} + \cdots
      | c = 
}}
{{eqn | r = \dfrac {\pi^4} {90}
      | c = 
}}
{{eqn | o = \approx
      | r = 1 \cdotp 08232 \, 3 \ldots
      | c = 
}}
{{end-eqn}}

Proof:
Create a multiplication table where the column down the {{LHS}} and the row across the top each contains the terms of zeta function of $2$:
:<nowiki>$\begin {array} {c|cccccccccc}
\paren {\map \zeta 2}^2 & \paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {2^2} } & \paren {\dfrac 1 {3^2} } & \paren {\dfrac 1 {4^2} } & \cdots \\
\hline
\paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {1^4} } & \paren {\dfrac 1 {1^2} } \paren {\dfrac 1 {2^2} } & \paren {\dfrac 1 {1^2} } \paren {\dfrac 1 {3^2} } & \paren {\dfrac 1 {1^2} } \paren {\dfrac 1 {4^2} } & \cdots \\
\paren {\dfrac 1 {2^2} }  & \paren {\dfrac 1 {2^2} } \paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {2^4} } & \paren {\dfrac 1 {2^2} } \paren {\dfrac 1 {3^2} } & \paren {\dfrac 1 {2^2} } \paren {\dfrac 1 {4^2} } & \cdots \\
\paren {\dfrac 1 {3^2} }  & \paren {\dfrac 1 {3^2} } \paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {3^2} } \paren {\dfrac 1 {2^2} } & \paren {\dfrac 1 {3^4} } & \paren {\dfrac 1 {3^2} } \paren {\dfrac 1 {4^2} } & \cdots \\
\paren {\dfrac 1 {4^2} }  & \paren {\dfrac 1 {4^2} } \paren {\dfrac 1 {1^2} } & \paren {\dfrac 1 {4^2} } \paren {\dfrac 1 {2^2} } & \paren {\dfrac 1 {4^2} } \paren {\dfrac 1 {3^2} } & \paren {\dfrac 1 {4^4} } & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\
\end {array}$</nowiki>
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:
{{begin-eqn}}
{{eqn | l = \paren {\map \zeta 2}^2
      | r = \paren {\sum_{i \mathop = 1}^\infty {\frac 1 {i^2} } } \paren {\sum_{j \mathop = 1}^\infty {\frac 1 {j^2} } }
      | c =
}}
{{eqn | r = \sum_{i \mathop = 1}^\infty \sum_{j \mathop = 1}^\infty {\frac 1 {i^2} } {\frac 1 {j^2} }
      | c = Product of Absolutely Convergent Series
}}
{{eqn | r = \sum_{i \mathop = 1}^{\infty} {\frac 1 {i^4} } + \sum_{i \mathop = 1}^\infty \sum_{j \mathop = {i + 1} }^\infty {\frac 1 {i^2} } {\frac 1 {j^2} } + \sum_{j \mathop = 1}^\infty \sum_{i \mathop = {j + 1} }^\infty {\frac 1 {i^2} } {\frac 1 {j^2} }
      | c = $\paren {i = j} + \paren {j > i} + \paren {j < i}$
}}
{{eqn | r = \map \zeta 4  + 2  \sum_{i \mathop = 1}^\infty \sum_{j \mathop = {i + 1} }^\infty {\frac 1 {i^2} } {\frac 1 {j^2} }
      | c = 
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = P_k
      | r = x \prod_{n \mathop = 1}^k \paren {1 - \frac {x^2} {n^2 \pi^2} }
      | c = 
}}
{{eqn | r = x \paren {1 - \dfrac {x^2} {1 \pi^2} } \paren {1 - \dfrac {x^2} {2^2 \pi^2} } \paren {1 - \dfrac {x^2} {3^2 \pi^2} } \cdots \paren {1 - \dfrac {x^2} {k^2 \pi^2} }
      | c = 
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = P_1
      | r = x - \frac {x^3} {\pi^2} \paren {\dfrac 1 {1^2} }
      | c = 
}}
{{eqn | l = P_2
      | r = x - \frac {x^3} {\pi^2} \paren {\dfrac 1 {1^2} +  \dfrac 1 {2^2} } + \frac {x^5} {\pi^4} \paren {\dfrac 1 {1^2} \dfrac 1 {2^2} }
      | c = 
}}
{{eqn | l = P_3
      | r = x - \frac {x^3} {\pi^2} \paren {\dfrac 1 {1^2} +  \dfrac 1 {2^2} + \dfrac 1 {3^2} } + \frac {x^5} {\pi^4} \paren {\dfrac 1 {1^2} \dfrac 1 {2^2} + \dfrac 1 {1^2} \dfrac 1 {3^2} + \dfrac 1 {2^2} \dfrac 1 {3^2} } - \frac {x^7} {\pi^6} \paren {\dfrac 1 {1^2} \dfrac 1 {2^2} \dfrac 1 {3^2} }
      | c = 
}}
{{eqn | l = P_4
      | r = x - \frac {x^3} {\pi^2} \paren {\dfrac 1 {1^2} +  \dfrac 1 {2^2} + \dfrac 1 {3^2} + \dfrac 1 {4^2} } + \frac {x^5} {\pi^4} \paren {\dfrac 1 {1^2} \dfrac 1 {2^2} + \dfrac 1 {1^2} \dfrac 1 {3^2} + \dfrac 1 {1^2} \dfrac 1 {4^2} + \dfrac 1 {2^2} \dfrac 1 {3^2} + \dfrac 1 {2^2} \dfrac 1 {4^2} + \dfrac 1 {3^2} \dfrac 1 {4^2} } - \frac {x^7} {\pi^6} \paren {\dfrac 1 {1^2} \dfrac 1 {2^2} \dfrac 1 {3^2} + \dfrac 1 {1^2} \dfrac 1 {2^2} \dfrac 1 {4^2} + \dfrac 1 {1^2} \dfrac 1 {3^2} \dfrac 1 {4^2} + \dfrac 1 {2^2} \dfrac 1 {3^2} \dfrac 1 {4^2} } + \frac {x^9} {\pi^8} \paren {\dfrac 1 {1^2} \dfrac 1 {2^2} \dfrac 1 {3^2} \dfrac 1 {4^2} }
      | c = 
}}
{{end-eqn}}
We make the following observations:
:$(1): \quad$ The number of terms added to calculate the coefficient of the $x^3$ term is $\dbinom k 1 = k$
:$(2): \quad$ The number of terms added to calculate the coefficient of the $x^5$ term is $\dbinom k 2$
:$(3): \quad$ For $k \ge 1$, the coefficient  of $x^3$ in $\ds P_k = - \dfrac 1 {\pi^2} \sum_{i \mathop = 1}^k \dfrac 1 {i^2}$
:$(4): \quad$ For $k \ge 2$, the coefficient  of $x^5$ in $\ds P_k = \dfrac 1 {\pi^4} \sum_{i \mathop = 1}^{k - 1} \sum_{j \mathop = {i + 1} }^k \paren {\frac 1 {i^2} } \paren {\frac 1 {j^2} } $
Expanding the product out to k, we get:
{{begin-eqn}}
{{eqn | l = P_k
      | r = x - \dfrac {x^3} {\pi^2} \sum_{i \mathop = 1}^k \dfrac 1 {i^2} + \frac {x^5} {\pi^4} \sum_{i \mathop = 1}^{k - 1} \sum_{j \mathop = {i + 1} }^k \paren {\frac 1 {i^2} } \paren {\frac 1 {j^2} } -  \cdots 
      | c = 
}}
{{end-eqn}}
Now recall the following two representations of the Sine of x:
{{begin-eqn}}
{{eqn | l = \sin x
      | r = x \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }
      | c = Euler Formula for Sine Function
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \sin x
      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \dfrac {x^3} {3!} + \dfrac {x^5} {5!} - \dfrac {x^7} {7!} + \cdots
      | c = Power Series Expansion for Sine Function
}}
{{end-eqn}}
Notice that by taking the limit of $P_k$ as $k \to \infty$, we obtain precisely the Euler Formula for Sine Function.
Equating the coefficient of $x^5$ in the Euler Formula for Sine Function with the Power Series Expansion for Sine Function, we have:
:$\ds \lim_{k \mathop \to \infty}  \dfrac 1 {\pi^4} \sum_{i \mathop = 1}^{k - 1} \sum_{j \mathop = {i + 1} }^k \paren {\frac 1 {i^2} } \paren {\frac 1 {j^2} } = \frac 1 {5!}$
Therefore:
:$\ds \sum_{i \mathop = 1}^\infty \sum_{j \mathop = {i + 1} }^\infty \paren {\frac 1 {i^2} } \paren {\frac 1 {j^2} } = \frac {\pi^4} {5!}$
Therefore:
{{begin-eqn}}
{{eqn | l = \paren {\map \zeta 2}^2 
      | r =  \map \zeta 4  + 2  \sum_{i \mathop = 1}^\infty \sum_{j \mathop = {i + 1} }^\infty {\frac 1 {i^2} } {\frac 1 {j^2} }
      | c = 
}}
{{eqn | l = \paren {\map \zeta 2}^2 
      | r =  \map \zeta 4 + 2  \dfrac {\pi^4} {5!}
      | c = 
}}
{{eqn | l = \map \zeta 4
      | r = \paren {\map \zeta 2}^2  - 2  \dfrac {\pi^4} {5!}
      | c = Rearranging
}}
{{eqn | r = \paren{\dfrac {\pi^2} 6 }^2 - \dfrac {\pi^4} {60}
      | c = Basel Problem
}}
{{eqn | r = \dfrac {\pi^4} {36} - \dfrac {\pi^4} {60}
      | c = 
}}
{{eqn | r = \dfrac {5\pi^4} {180} - \dfrac {3\pi^4} {180}
      | c = 
}}
{{eqn | r = \frac {2\pi^4} {180}
      | c = 
}}
{{eqn | r = \frac {\pi^4} {90}
      | c = 
}}
{{end-eqn}}
{{qed}}