If $\gcd(l,q)=1$, then $\sum_{p\equiv l}\frac{1}{p^s}=\frac{1}{\varphi(q)}\log(\frac{1}{s-1})+O(1)$ Suppose that $\gcd(l,q)=1$, then we claim that $$\sum_{p\equiv l}\frac{1}{p^s}=\frac{1}{\varphi(q)}\log(\frac{1}{s-1})+O(1)\quad\text{as $s\to1^+$}$$
I have already established the following facts that seem appropriate for proving this claim, they are as follows:
\begin{align*}
\sum_{p\equiv l}\frac{1}{p^s}&=\frac{1}{\varphi(q)}\sum_p\frac{\chi_0(p)}{p^s}+\frac{1}{\varphi(q)}\sum_{\chi\ne\chi_0}\overline{\chi(l)}\sum_p\frac{\chi(p)}{p^s}\quad(1)\\
&=\frac{1}{\varphi(q)}\sum_{p\nmid q}\frac{1}{p^s}+\frac{1}{\varphi(q)}\sum_{\chi\ne\chi_o}\overline{\chi(l)}\sum_p\frac{\chi(p)}{p^s}\quad(2)
\end{align*}
and
$$\sum_p\frac{1}{p^s}=\log(\frac{1}{s-1})+O(1)\quad \text{as $s\to 1^+$}\quad(3).$$

If I'm understanding the claim, (1) and (3) correctly,  we have that $$\sum_{p\equiv l}\frac{1}{p^s}=\frac{1}{\varphi(q)}\sum_p\frac{1}{p^s}+\frac{1}{\varphi(q)}\sum_{\chi\ne\chi_0}\overline{\chi(l)}\sum_p\frac{\chi(p)}{p^s}\quad(4)$$
since $gcd(l,q)=1$ and $p\equiv l\mod(q)$, so by definition of the trivial character $\chi_0(p)=1.$
Then by (3) it follows that 
$$\sum_{p\equiv l}\frac{1}{p^s}=\frac{1}{\varphi(q)}\big[\log(\frac{1}{s-1})+O(1)\big]+\frac{1}{\varphi(q)}\sum_{\chi\ne\chi_0}\overline{\chi(l)}\sum_p\frac{\chi(p)}{p^s}\quad(5).$$
I'm not sure where to go from here, or if the above is correct. In my class notes I have written that $\sum_{\chi\ne\chi_0}\overline{\chi(l)}$ is finite, but I'm not exactly sure why. 
Moreover, I have a theorem that says if $\chi$ is a non-trivial character, then $\sum_p\frac{\chi(p)}{p^s}$ remains bounded as $s\to 1^+$, since $\log(L(s,\chi))=\sum_p\frac{\chi(p)}{p^s}+O(1).$
With the above expression and the fact(?) that $\sum_{\chi\ne\chi_0}\overline{\chi(l)}$ is finite, I'm not sure if it's sufficient to say that the term on the right of (5) goes like $O(1)$ as $s\to 1^+$.
Could anyone point me in the right direction wrt proving this claim and clarify the point about $\sum_{\chi\ne\chi_0}\overline{\chi(l)}$ being finite?
 A: Let's look at
$$ \sum_{\chi \neq \chi_0} \overline{\chi(l)}.$$
The first question you should ask yourself is: what exactly is $\chi$? It is a Dirichlet character, but what sort of Dirichlet character? Closely related, why is your equation $(1)$ true, and where does $\varphi(q)$ come from?
A partial resolution to these questions comes from recognizing that $\chi$ refers do a Dirichlet character mod $q$. And the sum $\sum_{\chi \neq \chi_0}$ is over the non-trivial Dirichlet characters mod $q$. There are exactly $\varphi(q) - 1$ such characters (and a total of $\varphi(q)$ if you include the trivial character). Understanding and using this is key in proving your relation in $(1)$.
Now we return to looking at the sum
$$ \sum_{\chi \neq \chi_0} \overline{\chi(l)}.$$
Why is this finite? It is finite because there are at most $\varphi(q) - 1$ terms in the sum. And any finite sum of complex numbers is finite.
And you are correct, for $\chi \neq \chi_0$, it is true that
$$\sum_{p} \frac{\chi(p)}{p}$$
is finite for each $\chi$. Suppose we call $M$ the largest value (in magnitude) of this sum for the (finitely many) nontrivial $\chi$ mod $q$. Combined with the finiteness of the $\chi$ sum, we might bound
$$ \frac{1}{\varphi(q)} \sum_{\chi \neq \chi_0} \overline{\chi(l)} \sum_p \frac{\chi(p)}{p} \leq \frac{1}{\varphi(q)} (\varphi(q) - 1) \lvert M \rvert = O(1).$$
If everything else makes sense to you, this completes the ideas necessary to prove your initial claim.
