Riemann Zeta Function of 4/Proof 5

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Theorem

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

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


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:

\(\displaystyle \map \zeta 2\) \(=\) \(\displaystyle \paren {\frac 1 {1^2 } + \frac 1 {2^2 } + \frac 1 {3^2 } + \frac 1 {4^2 } + \cdots }\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots }\)


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

\(\displaystyle \displaystyle \paren {\map \zeta 2}^2\) \(=\) \(\displaystyle \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots } \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots }\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\sum_{i \mathop = 1}^{\infty} {\frac 1 {i^2 } } } \paren {\sum_{j \mathop = 1}^{\infty} {\frac 1 {j^2 } } }\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^{\infty} \sum_{j \mathop = 1}^{\infty} {\frac 1 {i^2 } } {\frac 1 {j^2 } }\) Product of Absolutely Convergent Series
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^{\infty} {\frac 1 {i^4 } } + \sum_{i \mathop = 2}^{\infty} \sum_{j \mathop = 1}^{i-1} {\frac 1 {i^2 } } {\frac 1 {j^2 } } + \sum_{j \mathop = 2}^{\infty} \sum_{i \mathop = 1}^{j - 1} {\frac 1 {i^2 } } {\frac 1 {j^2 } }\) $\paren {i = j } + \paren {j \lt i } + \paren {j \gt i }$
\(\displaystyle \) \(=\) \(\displaystyle \map \zeta 4 + 2 \dfrac {\pi^4} {5!}\) 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


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


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

$\blacksquare$