Let q be an odd prime and $\chi$ be the non-trivial real character modulo q. I am trying to prove that $L(1,\chi) = \sum_{n=1}^{\infty}\frac{\chi (n)}{n} > 0$.
Note: this question was first asked here: Summation of Legendre symbol but I have some additional questions based on my attempt at the solution.
There are multiple well-known proofs of the claim $L(1,\chi) \not = 0$. Now, based on that, the fact that $L(s,\chi)$ is representable as an Euler product of non-negative terms for $Re(s)>1$, and the fact that $L(s,\chi)$ is continuous at 1, one could conclude that $L(1,\chi) > 0$, right?
I was wondering if the above holds, and if there was a direct proof of the original claim, thereby proving the $L(1,\chi) \not = 0$ in a different way? Thanks.
There are several things to answer here. To be clear, we define $L(s, \chi)$ to by $\sum_n \frac{\chi(n)}{n^s}$ for $\mathrm{Re}(s)>1$ and extend $L(s, \chi)$ by analytic continuation to the rest of $\mathbb{C}$. Any product or sum on the variable $p$ is taken over prime values of $p$.
It is true, and not hard, that $\sum_n \frac{\chi(n)}{n} = L(1,\chi)$.
It is true, and not hard, that $\prod_{p} (1-\chi(p)/p^s)^{-1} = L(s, \chi)$ for $\mathrm{Re}(s)>1$.
It is true, but harder than showing $L(1, \chi) \neq 0$, that $\prod_{p} (1-\chi(p)/p)^{-1} = L(1, \chi)$.
It is by no means obvious that $\prod_{p} (1-\chi(p)/p)^{-1}$ is not $0$. For that matter, viewed as an Euler product, it isn't obvious that we don't get $\infty$.
Thus, due to issues 3 and 4, this is not a reasonable way to prove that $L(1, \chi) \neq 0$.
Proof of 1: Let $\chi$ be periodic with period $N$. Write the sum defining $L(s, \chi)$ as $\sum_{c=0}^{\infty} \sum_{d=1}^N \frac{\chi(cN+d)}{(cN+d)^s}$. The inner sum is $$\sum_{d=1}^N \frac{\chi(cN+d)}{(cN+d)^s} = \sum_{d=1}^N \chi(d) \left(\frac{1}{(cN)^s} + O \left( \frac{1}{c^{\mathrm{Re}(s)+1}} \right) \right) = O\left( \frac{1}{c^{\mathrm{Re}(s)+1}} \right) .$$ The second equality uses the fact that $\sum_{d=1}^N \chi(d)=0$. So the sum converges uniformly on $\mathrm{Re}(s) \geq a$ for any $a>0$. Therefore, the sum defines $L(s, \chi)$ for $\mathrm{Re}(s)>0$, not just $\mathrm{Re}(s)>1$.
Proof of 2: The sum and product converge absolutely for $\mathrm{Re}(s)>1$, so they may be rearranged at will.
Discussion of 3: As discussed in this MO answer, the equality $\prod_{p} (1-\chi(p)/p^a)^{-1} = L(a, \chi)$ for real $a$ is basically equivalent to showing $L(s, \chi) \neq 0$ for $\mathrm{Re}(s) \geq a$. It is known that $L(s,\chi)$ is nonzero on $\mathrm{Re}(s)=1$, so statement 3 is true, but this is obviously harder than showing $L(1, \chi) \neq 0$.
Discussion of 4: I don't see how you are going to show $\prod_{p} (1-\chi(p)/p)^{-1}$ is not $0$ or $\infty$ by looking at the product. For simplicity, let's talk about a quadratic character. Whenever $\chi(p)=1$, we get $(1-\chi(p)/p)^{-1} > \exp(1/p)$. If $\chi(p)$ is overwhelmingly $1$ between $N$ and $N^2$, then $$\prod_{N<p<N^2} (1-\chi(p)/p)^{-1} > \prod_{N<p<N^2} \exp(1/p) = \exp \left( \sum_{N<p<N^2} 1/p \right) \approx \exp(\log\log N^2 - \log \log N) = 2.$$
Similarly, if $\chi(p)$ is overwhelmingly $-1$ on $[N, N^2]$, then $\prod_{N<p<N^2} (1-\chi(p)/p)^{-1} < 1/2$. So, in order to show the product settles down to a nonzero finite number, you need to show that $\chi(p)$ does not overwhelmingly have one sign for primes in such intervals. Which is true! But one usually proves $L(1, \chi) \neq 0$ as a prelude to proving that $\chi(p)$ takes each sign infinitely often so, if you don't even know that yet, how can you hope to do this?