# Start at Basel Problem

In 18th century, Euler solved a well known problem, even Leibniz cannot solve it, either. This event made him become famous and confident. Basel problem shows $$\displaystyle \lim_{n \to +\infty} \left(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{n^2}\right)= \sum_{m=1}^{\infty} \dfrac{1}{m^2}=\dfrac{\pi^2}{6}$$. It is a wonderful theory, but I’m not satisfied with this. We can use it to find more interesting conclusions. \begin{align*} \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 \left(1-\dfrac{1}{2^2}\right) \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 \left(\dfrac{1}{8}-\dfrac{1}{24}\right)\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{align*} It’s true that Basel problem has an important relationship with Riemann $$\zeta$$ function, because Basel problem is in order to find $$\zeta(2)$$. About 100 years later, Riemann used one equation to build a bridge between functions and prime numbers. The equation is $$\displaystyle \zeta(s) =\sum_{n=1} ^{\infty} \dfrac{1}{n^s}= \prod_{p \text{ prime}} ^{\infty} (1-p^{-s})^{-1}$$. Here is Euler's proof. \begin{align*} \because \zeta(s) &=\dfrac{1}{1^s}+\dfrac{1}{2^s}+\dfrac{1}{3^s}+\cdots \\ \therefore \dfrac{1}{2^s} \zeta(s) &=\dfrac{1}{2^s}+\dfrac{1}{4^s}+\dfrac{1}{6^s}+\cdots \\ \therefore \left(1-\dfrac{1}{2^s}\right)\zeta(s) &=\dfrac{1}{1^s}+\dfrac{1}{3^s}+\dfrac{1}{5^s}+\cdots \\ \because \dfrac{1}{3^s} \times \left(1-\dfrac{1}{2^s}\right)\zeta(s) &=\dfrac{1}{3^s} \times \dfrac{1}{1^s}+\dfrac{1}{3^s} \times \dfrac{1}{3^s}+\dfrac{1}{3^s} \times \dfrac{1}{5^s}+\cdots \\ \therefore \left(1- \dfrac{1}{3^s}\right) \left(1-\dfrac{1}{2^s}\right)\zeta(s) &=\dfrac{1}{1^s}+\dfrac{1}{5^s}+\dfrac{1}{7^s}+\cdots \\ \therefore \left(1-\dfrac{1}{5^s}\right)\left(1- \dfrac{1}{3^s}\right)\left(1-\dfrac{1}{2^s}\right)\zeta(s) &=\dfrac{1}{1^s}+\dfrac{1}{7^s}+\dfrac{1}{11^s}+\cdots \\ \therefore 1 &=\zeta(s)\left(1-\dfrac{1}{2^s}\right)\left(1- \dfrac{1}{3^s}\right)\left(1-\dfrac{1}{5^s}\right)\cdots \\ \therefore \zeta(s) &=\dfrac{1}{\left(1-\dfrac{1}{2^s}\right)\left(1- \dfrac{1}{3^s}\right)\left(1-\dfrac{1}{5^s}\right)\left(1-\dfrac{1}{7^s}\right)\cdots} \\ \therefore \sum_{n=1} ^{\infty} \dfrac{1}{n^s} &= \prod_{p \text{ prime}} ^{\infty} (1-p^{-s})^{-1} \end{align*}

Note by Edward Christian
1 year, 2 months ago

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Are you actually 13?

- 9 months, 2 weeks ago

My friend, you’re just 12, too.

- 7 months ago