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# Lecture 24: The Riemann Zeta Function ## Euler's Product Formula For $s > 1$ we have $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{p \ prime} (1-p^{-s})^{-1}.$$ *Proof:* $$\begin{aligned} \prod_{p \ prime} (1-p^{-s})^{-1} &= \prod_{p \ prime} (1 + p^{-s} + p^{-2s} +...) \\ &= (1 + 2^{-...

# Lecture 24: The Riemann Zeta Function ## Euler's Product Formula For $s > 1$ we have $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{p \ prime} (1-p^{-s})^{-1}.$$ *Proof:* $$\begin{aligned} \prod_{p \ prime} (1-p^{-s})^{-1} &= \prod_{p \ prime} (1 + p^{-s} + p^{-2s} +...) \\ &= (1 + 2^{-s} + 2^{-2s} +...)(1 + 3^{-s} + 3^{-2s} +...)(1 + 5^{-s} + 5^{-2s} +...)... \\ &= \sum_{n=1}^{\infty} \frac{1}{n^s} = \zeta(s). \end{aligned}$$ ## Analytic Continuation of $\zeta(s)$ For $Re(s) > 1$ we have $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}.$$ We want to find a function that agrees with $\zeta(s)$ for $Re(s) > 1$ but is defined for $Re(s) \le 1$. Let $$f(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^s} = 1 - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} +...$$ Then $$\begin{aligned} f(s) &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^s} = \sum_{n=1}^{\infty} \frac{1}{n^s} - 2 \sum_{n=1}^{\infty} \frac{1}{(2n)^s} \\ &= \sum_{n=1}^{\infty} \frac{1}{n^s} - \frac{2}{2^s} \sum_{n=1}^{\infty} \frac{1}{n^s} \\ &= \zeta(s) - \frac{1}{2^{s-1}} \zeta(s) = (1 - 2^{1-s}) \zeta(s). \end{aligned}$$ So $$\zeta(s) = \frac{f(s)}{1 - 2^{1-s}}.$$ We know that $f(s)$ converges for $Re(s) > 0$. ## Example $$\begin{aligned} f(1) &= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} +... = \ln{2} \\ \zeta(1) &= \frac{f(1)}{1 - 2^{1-1}} = \frac{\ln{2}}{0} = \infty. \end{aligned}$$