Let $\chi$ is a non-principal character. Show that the sum $\sum_{p}\frac{\chi(p)}{p}$ is conditionally convergent. Then show that the product $\prod_{p}(1-\frac{\chi(p)}{p})^{-1}$ is conditionally convergent to $L(1,\chi)$.
Using summation by parts, I can show that $\sum_{n}\frac{\chi(n)}{n}$ is conditionally convergent. But it seems that the method cannot be applied to $\sum_{p}\frac{\chi(p)}{p}$ since $\frac{1}{n}1_{\mathcal{P}}(n)$ is not monotone.
On
Use partial summation to show that \[\sum_{p \leq x} \frac{\chi(p)}{p} = \frac{1}{x \log x} \sum_{p \leq x} \chi(p) \log p - \int_{2}^{x} \frac{\log t + 1}{t^2 (\log t)^2} \sum_{p \leq t} \chi(p) \log p \, dt.\] Now \[\sum_{p \leq x} \chi(p) \log p = \sum_{n \leq x} \chi(n) \Lambda(n) + O\left(\sqrt{x} \log x\right).\] Via the usual methods as for proving the prime number theorem, one can then show that \[\sum_{n \leq x} \chi(n) \Lambda(n) = O\left(x e^{-c\sqrt{\log x}}\right)\] if $\chi$ is nonprincipal. (Note that this is by far the hardest part; it's essentially equivalent to the prime number theorem in arithmetic progressions, which in turn is essentially equivalent to the nonvanishing of $L(s,\chi)$ along the line $\Re(s) = 1$. A standard reference for this estimate is chapter 11 of Montgomery and Vaughan.)
This implies the result, because then the first term in the earlier expression for $\sum_{p \leq x} \frac{\chi(p)}{p}$ is $o(1)$ and the second term is \[\int_{2}^{\infty} \frac{\log t + 1}{t^2 (\log t)^2} \sum_{p \leq t} \chi(p) \log p \, dt - \int_{x}^{\infty} \frac{\log t + 1}{t^2 (\log t)^2} \sum_{p \leq t} \chi(p) \log p \, dt,\] and the first integral is some finite number and the second is $o(1)$ due to the bound $\sum_{p \leq x} \chi(p) \log p = O\left(x e^{-c\sqrt{\log x}}\right)$.
EDIT: Note that for $\Re(s) > 1$, \[\log L(s,\chi) = \log \prod_p \frac{1}{1 - \chi(p) p^{-s}} = \sum_p \log \frac{1}{1 - \chi(p) p^{-s}} = \sum_p \sum_{k = 1}^{\infty} \frac{\chi(p)^k}{k p^{ks}}\] via the Euler product for $L(s,\chi)$ and the Taylor series expansion for $\log$. All of this is justified because $\Re(s) > 1$, so everything is absolutely convergent. It follows that for $\Re(s) > 1$, \[\sum_p \frac{\chi(p)}{p^s} = \log L(s,\chi) - \sum_p \sum_{k = 2}^{\infty} \frac{\chi(p)^k}{k p^{ks}}.\] The sum on the right-hand side is absolutely convergent for $\Re(s) > 1/2$, and so defines a holomorphic function on this right-half plane. Moreover, by the zero-free region for $L(s,\chi)$, $\log L(s,\chi)$ defines a holomorphic function in the same zero-free region. It follows by analytic continuation that the left-hand side extends to a holomorphic function in this same zero-free region.
In spite of all of this, one cannot conclude immediately this holomorphic function is equal to $\sum_p \frac{\chi(p)}{p}$ when $s = 1$, and so this sum converges conditionally. It does show however that the limit $\lim_{s \to 1^+} \sum_p \frac{\chi(p)}{p^s}$ exists.
On
We can use the following estimate which only needs the fact that $L(1,\chi)$ doesn't vanish for non-principal character: $$\sum_{n\leq x}\frac{\Lambda(n)\chi(n)}{n}=O_{\chi}(1)\quad (1) $$ for any $x\geq 1$ and non-principal character $\chi$ (see Terence Tao's blog Theorem 70).
Note that $$\sum_{n\leq x}\frac{\Lambda (n)\chi(n)}{n}=\sum_{p\leq x}\frac{\log (p)\chi(p)}{p}+\sum_{j=2}^\infty\sum_{p\leq x^{1/j}}\frac{\log (p)\chi(p^j)}{p^j}.$$ We crudely bound $\frac{\log p\chi(p^j)}{p^j}\ll \frac{1}{2^{j/2}}\frac{\log p}{p^{3/2}}$ for $j\geq 2$. Since $\sum_{n}\frac{\log n}{n^{3/2}}=O(1)$, we obtain $$\sum_{n\leq x}\frac{\Lambda (n)\chi(n)}{n}=\sum_{p\leq x}\frac{\log p\chi(p)}{p}+O_\chi(1).$$ It follows from $(1)$ that $$\sum_{n\leq x}\frac{\log p\chi(p)}{p}=O_{\chi}(1).$$ Now we use summation by parts to obtain $$\sum_{x\leq p\leq y}\frac{\chi(p)}{p}=\sum_{x\leq p\leq y}\frac{\log p\chi(p)}{p}\frac{1}{\log y}-\int_{x}^y\sum_{x\leq y\leq t}\frac{\log p\chi(p)}{p}-\frac{1}{t\log^2 t}\,dt.$$ Since $\sum_{x\leq p\leq t}\frac{\log p\chi(p)}{p}=O(1)$ for all $x\leq t\leq y$, we have $$\sum_{x\leq p\leq y}\frac{\chi(p)}{p}=O(\frac{1}{\log y}+\frac{1}{\log x}),$$ here I omit $\chi$ in the big O. We see from Cauchy's criterion that $\sum_{p}\frac{\chi(p)}{p}$ is convergent.
To show that $\prod_{p}(1-\frac{\chi(p}{p})^{-1}$ is conditionally convergent to $L(1,\chi)$. Taking logarithm, it suffices to show that $$\log L(1,\chi)=-\sum_{p}\log(1-\frac{\chi(p)}{p}).$$ Since $(1,\chi)\neq 0$, we have $\log(s,\chi)$ is holomprhic on a neighborhood of $s=1$, it follows that $\log L(1,\chi)=\lim_{s\to 1^+}\log L(s,\chi)$. Thus it suffices to show $$\lim_{s\to 1^+}\log L(s,\chi)=-\sum_{p}\log (1-\frac{\chi(p)}{p}).$$ Note that \begin{align}\log L(s,\chi)&=\sum_{p}\frac{\chi(p)}{p^s}+\sum_{j=2}^\infty\sum_p\frac{\chi(p)^j}{jp^{sj}}\\ &=\sum_{p}\frac{\chi(p)}{p^s}+\text{something continuous on }\mathrm{Re}(s)> \frac{1}{2} \end{align} for $s>1$ and $$-\log(1-\frac{\chi(p)}{p})=\sum_p\frac{\chi(p)}{p}+\sum_{j=2}^\infty\sum_p\frac{\chi(p)^j}{jp^j}.$$ thus it suffices to show that $$\lim_{s\to 1^+}\sum_{p}\frac{\chi(p)}{p^s}=\sum_{p}\frac{\chi(p)}{p}.$$ To show this, we mimic the proof of Abel's theorem. By partial summation, we have \begin{align} \sum_{n=1}^N \left(\frac{\chi(p_n)}{p_n^s}-\frac{\chi(p_n)}{p_n}\right)&=\sum_{n=1}^N \frac{\chi(p_n)}{p_n}\left(\frac{1}{p_n^{s-1}}-1\right)\\ &=\sum_{n=1}^N \frac{\chi(p_n)}{p_n}\left(\frac{1}{p_N^{s-1}}-1\right)+\sum_{n=1}^{N-1}\left(\sum_{k=1}^n\frac{\chi(p_k)}{p_k}\right)\left(\frac{1}{p_n^{s-1}}-\frac{1}{p_{n+1}^{s-1}}\right). \end{align} Sending $N$ to infinity, we obtain \begin{align}\sum_p \frac{\chi(p)}{p^s}-\sum_p\frac{\chi(p)}{p}&=-\sum_{n=1}^\infty \frac{\chi(p_n)}{p_n}+\sum_{n=1}^{\infty}\left(\sum_{k=1}^n\frac{\chi(p_k)}{p_k}\right)\left(\frac{1}{p_n^{s-1}}-\frac{1}{p_{n+1}^{s-1}}\right)\\ =&-\left(1-\frac{1}{2^{s-1}}\right)\sum_{n=1}^\infty\frac{\chi(p_n)}{p_n}-\sum_{n=1}^{\infty}\left(\sum_{k=n+1}^\infty\frac{\chi(p_k)}{p_k}\right)\left(\frac{1}{p_n^{s-1}}-\frac{1}{p_{n+1}^{s-1}}\right). \end{align} Since $\sum_{n=1}^\infty\frac{\chi(p_n)}{p_n}$ is convergent, for every $\varepsilon>0$, we can choose $n$ sufficiently large (say $n\geq N$ for some $N$) such that $|\sum_{n+1}^\infty\frac{\chi(p_k)}{p_k}|\leq\varepsilon$ for all $n\geq N$, thus we have $$\left|\sum_{n\geq N}\left(\sum_{k=n+1}^\infty\frac{\chi(p_k)}{p_k}\right)\left(\frac{1}{p_n^{s-1}}-\frac{1}{p_{n+1}^{s-1}}\right)\right|\leq\varepsilon\sum_{n\geq N}\left(\frac{1}{p_n^{s-1}}-\frac{1}{p_{n+1}^{s-1}}\right)\leq\varepsilon.$$ It remains to control $-(1-\frac{1}{2^{s-1}})\sum_{n=1}^\infty\frac{\chi(p_n)}{p_n}$ and $\sum_{n< N}(\sum_{k=n+1}^\infty\frac{\chi(p_k)}{p_k})(\frac{1}{p_n^{s-1}}-\frac{1}{p_{n+1}^{s-1}})$, these two functions are continuous at $s=1$, thus we can choose $s$ sufficiently close to $1$ to make them less than $\varepsilon$. The claim then follows.
complicated. there is a proof sketch when $\chi$ is a non-principal non-quadratic character (i.e. when it is a complex valued character) :
the main trick is to use that $\frac{1}{\chi(q)}\sum_{\chi \pmod q} \chi(n) = \delta_{\ n \ \equiv\ 1 \pmod q}$ (the discrete Fourier transform of size $\phi(q)$)
and that $L(s,\chi)$ is holomorphic (on $ Re(s) > 0$ is enough and easy to prove) whenever $\chi$ is not the principal character $\chi_0$ (i.e. $\chi_0(n)=1$ whenever $gcd(n,q)$).
hence if $\chi \ne \chi_0$ and $\lim_{s \to 1^+} \sum_p \chi(p)/p^s$ diverges then $L(s,\chi)$ has a $0$ at $s=1$,
but if it is a non real character then the character comes with its complex conjugate $\overline{\chi}$ which in that case also has a zero at $s=1$, and hence two terms of the product $$F_q(s) = \prod_{\chi \pmod q} L(s,\chi)$$ would have a zero at s=1, but only $L(s,\chi_0)$ has a pole at $s=1$, of order $1$, hence $F_q(s)$ has a zero at $s=1$, a contradiction since : $$\ln F_q(s) = \phi(q) \sum_{p^k \equiv 1 \pmod q} \frac{1}{k p^{k s}}$$ which diverges to $+\infty$ when $s \to 1^+$.
then prove that $L(s,\chi)$ doesn't have any zero on $Re(s) =1$, using the same argument as for $\zeta(s)$ : https://en.wikipedia.org/wiki/Prime_number_theorem#Proof_sketch
which is based on the same kind of trick : that $\zeta(s)$ is meromorphic, that it doesn't have any pole on $Re(s) = 1, s \ne 1$, and that $\ln \zeta(s)$ is a Dirichlet series with non-negative coefficients, hence using $$\zeta(x)^3\zeta(x+iy)^4\zeta(x+2iy)|=\exp\sum_{n,p}\frac{3+4\cos(ny\log p) +\cos (2ny\log p)}{np^{nx}}\ge 1$$ if it had a zero at $1+it$ it would have a pole at $1+2it$, a contradiction.