This question expands on this one from MSE.

In the literature about Dirichlet $L$-series, I found that their Euler products:

$$L(s, \chi) =\prod_p \bigg(\frac {1}{1-\frac{\chi(p)}{p^s}} \bigg)$$

are typically considered to be only converging for $\Re(s)>1$.

However, there seems to be an exception to this rule since Euler proved that:

$$L(1, \chi_4) =\prod_p \bigg(\frac {p}{p-\chi_4(p)} \bigg)=\prod_p \bigg(\frac {p}{p-\sin\left(\frac{p \,\pi}{2}\right)} \bigg)=\frac{3}{4}\cdot\frac{5}{4}\cdot\frac{7}{8}\cdot\frac{11}{12}\cdot\frac{13}{12}\dots=\beta(1)=\frac{\pi}{4}$$

does converge (albeit slowly).

I then decided to explore values for $\Re(s) \lt 1$ and numerical evidence suggests that the Euler-product:

$$\prod_p \bigg(\frac {p^s}{p^s-\sin\left(\frac{p \,\pi}{2}\right)} \bigg)$$

also (slowly) converges in the domain $\frac12 < \Re(s) \le 1$.

**Questions:**

1) Is the Euler product for $L(s,\chi_4)$ the only one known to converge for $s=1$?

2) Does this particular Euler product indeed converge in the right half of the strip?

Thanks.

ABCDvevewrote similarly in his/her answer.) $\endgroup$ – Noam D. Elkies Jun 17 '15 at 21:40provethe Euler product converges at any $s$ with ${\rm Re}(s) < 1$. $\endgroup$ – KConrad Jun 17 '15 at 23:14