# Riemann's $\zeta$ Function and Prime Numbers

We know that \begin{aligned} \because \dfrac{\pi^2}{6} &=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots \\ \therefore \dfrac{1}{2^2} \times \dfrac{\pi^2}{6} &=\dfrac{1}{2^2} \times \dfrac{1}{1^2}+\dfrac{1}{2^2} \times \dfrac{1}{2^2}+\dfrac{1}{2^2} \times \dfrac{1}{3^2}+\cdots \\ \therefore \dfrac{1}{2^2} \times \dfrac{\pi^2}{6} &=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\cdots \\ \therefore (1-\dfrac{1}{2^2})\dfrac{\pi^2}{6} &=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots \\ \dfrac{3}{4} \times \dfrac{\pi^2}{6} &=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots \\ \therefore \dfrac{\pi^2}{8} &=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots \\ \dfrac{\pi^2}{24} &=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\cdots \\ \therefore (\dfrac{1}{8}-\dfrac{1}{24})\pi^2 &=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\cdots \\ \therefore \dfrac{\pi^2}{12} &=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\cdots \\ \end{aligned} As for $\dfrac{\pi^2}{6}=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots$, let $z$ takes $2$'s place, we can get the Riemann's $\zeta$ function. And if we continue, we can get \begin{aligned} \zeta(z) &=\dfrac{1}{1^z}+\dfrac{1}{2^z}+\dfrac{1}{3^z}+\cdots \\ (1-\dfrac{1}{2^z})\zeta(z) &=\dfrac{1}{1^z}+\dfrac{1}{3^z}+\dfrac{1}{5^z}+\cdots \\ \therefore \dfrac{1}{3^z} \times (1-\dfrac{1}{2^2})\zeta(z) &=\dfrac{1}{3^z} \times \dfrac{1}{1^z}+\dfrac{1}{3^z} \times \dfrac{1}{3^z}+\dfrac{1}{3^z} \times \dfrac{1}{5^z}+\cdots \\ \therefore (1- \dfrac{1}{3^z}) (1-\dfrac{1}{2^z})\zeta(z) &=\dfrac{1}{1^z}+\dfrac{1}{5^z}+\dfrac{1}{7^z}+\cdots \\ \therefore (1-\dfrac{1}{5^z})(1- \dfrac{1}{3^z}) (1-\dfrac{1}{2^z})\zeta(z) &=\dfrac{1}{1^z}+\dfrac{1}{7^z}+\dfrac{1}{11^z}+\cdots \\ \therefore 1 &=\zeta(z)(1-\dfrac{1}{2^z})(1- \dfrac{1}{3^z})(1-\dfrac{1}{5^z})\cdots \\ \therefore \zeta(z) &=\dfrac{1}{(1-\dfrac{1}{2^z})(1- \dfrac{1}{3^z})(1-\dfrac{1}{5^z})(1-\dfrac{1}{7^z})\cdots} \end{aligned} Note by Edward Christian
4 weeks ago

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