The sum of the reciprocals of the primes diverges, $\sum_p\frac{1}{p}=\infty$. This can be seen using the Euler product for the Riemann zeta function (say $s>1$ real), $\zeta(s)=\prod_p\frac{1}{1-p^{-s}}$,
$$
\log\left(\zeta(s)\right)=\sum_p-\log(1-p^{-s})=\sum_{p,n}\frac{p^{-ns}}{n}=\sum_p\frac{1}{p^s}+O(1),
$$
and letting $s\to1^+$.
Dirichlet extends this by replacing $\zeta(s)$ with $L(s,\chi)=\prod_p\frac{1}{1-\chi(p)p^{-s}}$,
$$
\log(L(s,\chi))=-\sum_p\log(1-\chi(p)p^{-s})=\sum_{n,p}\frac{\chi(p)}{np^{ns}}=\sum_p\frac{\chi(p)}{p^s}+O(1),
$$
and using properties of characters to select the progression (i.e. average over all characters and "shift" $1\bmod q$ to $a\bmod q$)
$$
\frac{1}{\phi(q)}\sum_{\chi}\bar{\chi}(a)\log(L(s,\chi))=\sum_pp^{-s}\sum_{\chi}\chi(pa^{-1})+O(1)=\sum_{p\equiv a(q)}\frac{1}{p^s}+O(1).
$$
The proof goes through (letting $s\to1^+$) as long as $L(1,\chi)\neq0$ for all $\chi$. There is a pole for the principal character $\chi_0$, and the other $L(1,\chi)$ values are finite, but we get "$\infty-\infty$" on the LHS if some of these values are zero.
The "relate arithmetic progressions to characters" question is basically orthogonality (some algebra)
$$
\begin{align*}
\sum_{\chi}\chi(a)&=
\left\{
\begin{array}{cc}
1 & \chi=\chi_0\\
0 & \text{else},\\
\end{array}
\right.
\\
\sum_{a\in(\mathbb{Z}/q\mathbb{Z})^{\times}}\chi(a)&=
\left\{
\begin{array}{cc}
1 & a\equiv1(q)\\
0 & \text{else}.\\
\end{array}
\right.
\end{align*}
$$
If you want to see some proofs, try Davenport's Multiplicative Number Theory or these somewhat concise notes I wrote once upon a time (with a few proofs of non-vanishing of $L(1,\chi)$).
To more directly address the concerns of the question, the algebraic trick to select an arithmetic progression using characters is essentially the sum
$$
\frac{1}{\phi(q)}\sum_{\chi\bmod q}\chi(pa^{-1})=
\left\{
\begin{array}{cc}
1 & p\equiv a\bmod q\\
0 & \text{else},\\
\end{array}
\right.
$$
which is a direct consequence of the orthogonality relations above.
For example, lets consider $q=3$. There are two characters modulo 3,
$$
\chi_0 = (0,1,1)=(\chi_0(0), \chi_0(1),\chi_0(2)), \quad \psi = (0,1,-1)=(\psi(0), \psi(1),\psi(2)),
$$
(listing the values on $0,1,2 \bmod 3$). If you average these, they "interfere" to give
$$
\frac{1}{\phi(q)}\sum_{\chi\bmod q}\chi=\frac{1}{2}(0,2,0)=(0,1,0)
$$
which picks out the residue class $1\bmod 3$. To pick out the residue class $2\bmod 3$ we can "shift the indices" by considering the translated characters ($x\mapsto 2^{-1}x=2x$ working $\bmod 3$)
$$
\chi_0(2x) = (0,1,1)=(\chi_0(0), \chi_0(2),\chi_0(4)), \quad \psi(2x) = (0,-1,1) =(\psi(0), \psi(2),\psi(4)).
$$
Averaging these instead, and noting $\chi(2x)=\chi(2)\chi(x)$, we get
$$
\frac{1}{\phi(q)}\sum_{\chi\bmod q}\overline{\chi}(a)\chi=\frac{1}{2}(1\cdot(0,1,1)+(-1)\cdot(0,1,-1))=(0,0,1),
$$
which picks out $2\bmod 3$.
[Note that $2\cdot2=4\equiv1\bmod 3$, i.e. $2=2^{-1}\bmod 3$, and $\chi(a^{-1})=\overline{\chi}(a)$ by multiplicativity if the bar or $a^{-1}$ is confusing.]
Best Answer
Sure. By Dirichlet, we can find infinitely many primes $p_i\equiv 1\pmod b$ and infinitely many primes $q_i\equiv a\pmod b$, in which case $p_i\times q_i$ works.