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Let $\chi$ be a Dirichlet character $\bmod q$. We have $$\sum_{n=0}^{\infty} (-1)^{n-1} \chi(n) n^{-s}=\prod_p \left(1-(-1)^{p-1}\frac{\chi(p)}{p^s}\right)^{-1}=(1+\chi(2)2^{-s})^{-1}\prod_{p>2}\left(1-\frac{\chi(p)}{p^s}\right)^{-1}$$$$=(1+\chi(2)2^{-s})^{-1}(1-\chi(2)2^{-s})L(s,\chi)$$ Using the Euler product, how do I show that $$(1+\chi(2)2^{-s})^{-1}(1-\chi(2)2^{-s})=1-\chi(2)2^{1-s}$$

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2 Answers 2

up vote 5 down vote accepted

Your Euler product factorization is incorrect. It should proceed like the following: $$\sum_{n=1}^\infty\frac{(-1)^{n-1}\chi(n)}{n^s}=\left(1-\frac{2\chi(2)}{2^s}\right)\sum_{n=1}^\infty\frac{\chi(n)}{n^s}=\frac{1-\chi(2)2^{1-s}}{1-\chi(2)2^{-s}}\prod_{p>2}\left(\frac{1}{1-\chi(p)p^{-s}}\right),$$ or simply $(1-\chi(2)2^{1-s})L(s,\chi)$. Compare to equations $(16)-(17)-(18)-(19)$ on Mathworld.

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You don't, since it's false. Say $\chi(2)=1$, $s=0$, then you are trying to prove $$(1+1)^{-1}(1-1)=1-2$$

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