TITLE: Growth of $\zeta_{\mathbf Q[\cos(\frac{\pi}{2^{n+1}})]}(2)$
QUESTION [14 upvotes]: Let $K_n$ be the field $\mathbf Q[\cos(\frac{\pi}{2^{n+1}})]$ (the real subfield of the cyclotomic field $\mathbf Q[e^{\frac{i\pi}{2^{n+1}}}]$). 
Is there anything known about the growth of the special values of the Dirichlet $\zeta$ function 
$$\zeta_{K_n}(2)=\prod_{\mathfrak p}\frac{1}{1-\frac{1}{N(\mathfrak p)^2}}$$ when $n\to \infty$ ?
The best I could do was to use the inequalities
$$\left(1-\frac{1}{p^2}\right)^{ab}\leq \left(1-\frac{1}{p^{2a}}\right)^{b}\leq \left(1-\frac{1}{p^{2ab}}\right)$$
(that are verified in our range) to obtain the inequalities
$$2\ \left(\frac{\pi^2}{3}\right)^{2^n}\ =\ 2^{1-2^n}\ \zeta(2)^{2^n}\geq \zeta_{K_n}(2)\geq 2\ \left(1-\frac{1}{2^{n+1}}\right)\ \zeta(2^{n+1})$$
that are quite unilluminating since the left hand side tends to infinity exponentially fast, while the right hand side converges to 2. The only thing we use here is that $2$ is totally ramified and that the other primes are not ramified. I guess it is possible to use more of our knowledge of the arithmetic in $K_n$ to obtain a precise idea of the behaviour of the sequence $(\zeta_{K_n}(2))_n$ ... (maybe by a standard use of the Chebotarev theorem - I'm quite ignorant in this kind of questions). 
I'm almost certain that $\zeta_{K_n}(2)$ tends to infinity exponentially fast (and that's what I'd like to check).

REPLY [18 votes]: Actually $\zeta_{K_n}(\sigma)$ is bounded for any fixed $\sigma > 1$.
Let $N = 2^n = [K_n : {\bf Q}]$.  Then all the local factors of 
$\zeta(\sigma)$, other than the factor $(1-2^{-\sigma})^{-1}$ for
the prime above $2$, are of the form $(1 - q^{-\sigma})^{-g}$,
where $q$ is a prime power congruent to $\pm 1 \bmod 4N$, and $g \mid N$.
Thus 
$$
\zeta_{K_n}(\sigma) < \frac1{1-2^{-\sigma}} 
  \prod_{m=1}^\infty \frac1{(1 - (2mN)^{-\sigma})^N}
$$
and the product approaches $1$ as $n \to \infty$ because
its logarithm behaves like
$\sum_{m=1}^\infty N/(2mN)^\sigma = \zeta(\sigma) \, / \, 2^\sigma N^{\sigma-1} \to 0$.
Since each local factor exceeds $1$, it follows that in fact
$\zeta_{K_n}(\sigma) \to (1-2^{-\sigma})^{-1}$ as $n \to \infty$.