Showing a limit related to Riemann Zeta Function is finite and nonzero (number theory) 
Show that $$\lim_{s\to 1^+} (s-1) \prod_{q \equiv 1 \pmod 4} (1-q^{-s})^{-1}$$ exists and is a finite nonzero real number.

I showed that $$\lim_{s\to 1^+} \prod_{q \equiv 1 \pmod 4} (1-q^{-s})^{-1} = \infty.$$
Any ideas on how to proceed/conclude? (I am using Leveque's Number Theory.)
 A: I can see two reasonable interpretations of the product:


*

*$q$ runs over all primes $\equiv 1 \pmod{4}$, and

*$q$ runs over all positive integers $\equiv 1 \pmod{4}$; but then we would have a problem for $q = 1$, so it ought to be the product over all $q > 1$ that are $\equiv 1 \pmod{4}$. That makes the reasonability of this interpretation tenuous. Still, let's keep it.


The claim is wrong for both of these interpretations.
First interpretation: Let $\chi$ be the nonprincipal character modulo $4$, i.e.
$$\chi(n) = \begin{cases}\; 1 &\text{if } n \equiv 1 \pmod{4} \\ -1 &\text{if } n \equiv 3 \pmod{4} \\ \;0 &\text{if } n \equiv 0 \pmod{2}\end{cases}$$
and $L(s,\chi)$ the associated $L$-function. Further define
$$F_1(s) = \prod_{q \equiv 1 \pmod{4}} \bigl(1 - q^{-s}\bigr)^{-1} \qquad\text{and} \qquad F_3(s) = \prod_{q\equiv 3 \pmod{4}} \bigl(1 - q^{-s}\bigr)^{-1}$$
for $\operatorname{Re} s > 1$. From the Euler product of $L(s,\chi)$ we immediately obtain
\begin{align}
F_1(s) &= L(s,\chi)\prod_{q \equiv 3 \pmod{4}}\bigl(1 + q^{-s}\bigr) \\
&= L(s,\chi) \prod_{q \equiv 3 \pmod{4}} \frac{1 - q^{-2s}}{1- q^{-s}} \\
&= L(s,\chi)\cdot\frac{F_3(s)}{F_3(2s)}
\end{align}
and hence, using the Euler product of the Riemann zeta function,
$$F_1(s)^2 = \bigl(F_1(s)F_3(s)\bigr) \cdot \frac{L(s,\chi)}{F_3(2s)} = \zeta(s)\cdot \frac{\bigl(1 - 2^{-s}\bigr)L(s,\chi)}{F_3(2s)} \tag{1}$$
for $\operatorname{Re} s > 1$. In the fraction on the right hand side, $\bigl(1 - 2^{-s}\bigr)$ is an entire function that attains the value $\frac{1}{2}$ at $s = 1$. $L(s,\chi)$ is given by a Dirichlet series that converges for $\operatorname{Re} s > 0$ (it is in fact an entire function too, but we don't need that), and its value at $s = 1$ is $\frac{\pi}{4}$. The denominator $F_3(2s)$ is holomorphic and zero-free for $\operatorname{Re} s > \frac{1}{2}$ (by the Euler product for $F_3$), it is real and $> 1$ on $\bigl(\frac{1}{2},+\infty\bigr)$. Using the known behaviour of $\zeta$, we consequently have
$$\lim_{s \to 1^+} (s-1)F_1(s)^2 = \frac{\pi}{8F_3(2)}\cdot \lim_{s \to 1^+} (s-1)\zeta(s) = \frac{\pi}{8F_3(2)} \in (0,+\infty).$$
Taking square roots, we find that
$$\lim_{s \to 1^+} \sqrt{s-1}\cdot F_1(s)$$
is a nonzero (strictly positive) real number, so
$$\lim_{s \to 1^+} (s-1)F_1(s) = 0.$$
Second interpretation: The product is absolutely convergent for $\operatorname{Re} s > 1$, and it's real and positive for real $s > 1$. Taking logarithms - in each case using the branch that is real on $(0,+\infty)$ - we obtain
\begin{align}
\log \prod_{q \equiv 1 \pmod{4}} \bigl(1 - q^{-s}\bigr)^{-1} &= - \sum_{q \equiv 1 \pmod{4}} \log \biggl(1 - \frac{1}{q^s}\biggr) \\
&= \underbrace{\sum_{q \equiv 1 \pmod{4}} \frac{1}{q^s}}_{F(s)} - \underbrace{\sum_{q \equiv 1 \pmod{4}} \Biggl(\frac{1}{q^s} + \log \biggl(1 - \frac{1}{q^s}\biggr)\Biggr)}_{G(s)}\,.
\end{align}
Using $\lvert x + \log (1-x)\rvert < x^2$ for $0 < x < \frac{1}{2}$ we see that the series defining $G$ converges absolutely for $\operatorname{Re} s > \frac{1}{2}$ and defines a holomorphic function there. It is easy to see that $F$ has a meromorphic continuation to the half-plane $\operatorname{Re} s > 0$, and that continuation is holomorphic except for a simple pole with residue $\frac{1}{4}$ at $s = 1$. Thus there is an $\varepsilon > 0$ such that
$$\prod_{q \equiv 1 \pmod{4}} \bigl(1 - q^{-s}\bigr)^{-1} > \exp \frac{1}{8(s-1)}$$
for $1 < s < 1 + \varepsilon$. It follows that
$$\lim_{s \to 1^+} (s-1)\prod_{q \equiv 1 \pmod{4}} \bigl(1 - q^{-s}\bigr)^{-1} = +\infty.$$
