Hint on proof of convergence of Dedekind zeta function The following is an exercise in a number theory book (by Childress):
Let $K/\mathbb{Q}$ be a number field. Show that there series below representing $\zeta_K(s)$ 
$$\sum\limits_{n=1}^\infty \frac{\gamma_n}{n^s}\quad \text{where. } \gamma_n=\#\{\text{ideals }\mathfrak{a} \subseteq \mathcal{O}_K: N\mathfrak{a}=n\} $$ is absolutely convergent for $\Re(s)>1$. Are  the $\gamma_n$ bounded?
Here’s my attempted write up,
I can show by example that $\gamma_n$ need not be bounded: Take $K=\mathbb{Q}[i]$, and recall that any prime $p\equiv 1 \bmod 4$ splits $p=\mathfrak{p}_1\mathfrak{p}_2$ in $\mathcal{O}_K$. So $g=2, e=1,f=1$ and $N(\mathfrak{p}_i)=p$. Then $N(\mathfrak{p}_1^i\mathfrak{p}_2^{m-i})=p^m$ for $i=0, \ldots m$, and therefore $\gamma_{p^m} = m+1$. Hence letting $m\to \infty$, we see that $\gamma_{p^m}$ is not bounded.
I’m stuck on the initial problem of absolute convergence and the case of a general field $K$. Let $d=[K:\mathbb{Q}]$. The ``worst case'' scenario (i.e. bound on $\gamma_n$) is where all primes $p$ split completely $p=\mathfrak{p}_1\cdots \mathfrak{p}_d$. Then $\gamma_{p^m}$ is the number of solutions $(n_1, \ldots, n_d)$ in nonnegative integers to $n_1 + \cdots n_d =m$, and this is $\binom{m}{d-1}$. Note that $m=\log_p p^m = \frac{\log p^m}{\log p}$ so for $p\geq 3$ we have $m \leq \log(p^m)$. Also, $\binom{m}{d-1} \leq m^{d-1}$. Therefore 
    $$\gamma_{p^m} \leq (\log p^m)^{d-1}$$
Now let $n=p_1^{m_1} \cdots p_r^{m_r}$ be the prime factorization of an positive integer $n \in \mathbb{Z}$. Then 
$$ \gamma_n = \prod_{i=1}^r \gamma_{p_i^{m_i}} \leq \prod (\log p_i^{m_i})^{d-1} $$
By the AM-GM inequality we have 
$$ \prod \log p_i^{m_i} \leq \left( \frac{\sum_i \log p_i^{m_i}}{r} = \right)^r = \left(\frac{\log n}{r} \right) ^r $$
where $r$ is the number of primes dividing $n$.
So our bound on $\gamma_n$ is 
$$ \gamma_n \leq \left(\frac{\log n}{r} \right)^{r(d-1)}$$
A calculus exercise shows that $f(r)=(C/r)^r$ is maximized at $r=C/e$ (where $C$ is a constant). Hence  
$$ \gamma_n \leq e^{(\log n)(d-1)/e}$$
Hmm, it seems our bound has turned into an exponential function, so that is not good.
 A: The right way to think about this is by considering the Euler product.
The $p$-part of this is
$$\prod_{i}\left(1-\frac1{p^{a_is}}\right)^{-1}
=\prod_{i}\left(1+\frac1{p^{a_is}}+\frac1{p^{2a_is}}+\cdots\right)$$
where the $p^{a_i}$ are the norms of the primes in $\mathcal{O}_K$ dividing
$p$. As there are at most $d$ of these, then for $s>1$
$$1+\frac1{p^{a_is}}+\frac1{p^{2a_is}}+\cdots
\le
1+\frac1{p^{s}}+\frac1{p^{2s}}+\cdots=\left(1-\frac1{p^s}\right)^{-1}$$
and so
$$\prod_{i}\left(1+\frac1{p^{a_is}}+\frac1{p^{2a_is}}+\cdots\right)
\le\left(1-\frac1{p^s}\right)^{-d}.$$
For $s>1$ then $\zeta_K(s)\le\zeta(s)^d$.
This inequality holds for the coefficients in the Dirichlet series.
The coefficient of $1/n^s$ in the Dirichlet series of $\zeta(s)^d$
is $c_d(n)$ where this denotes the number of $d$-tuples $(n_1,\ldots,n_d)$
of positive integers with product $n$. You should be able to get this
from your formula for $\gamma_{p^m}$. Once you have the inequality
$\gamma_n\le c_d(n)$, then you win, since even if you didn't know the
Euler product then it's elementary that $\zeta(s)^d=\sum_n c_d(n)/n^s$.
I suspect your estimates on the $\gamma_n$ are a bit too loose to get
a manifestly convergent series.
