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I am following Marcus approach to determine the class number formula. It is on page 192, chapter 7. There is a first step that I do not get but it should very easy so it is frustrating. Probably I do not know some property of complex analysis.

He writes $\zeta_K(s)=\sum_{n=1}^{\infty}(j_n-hk)/(n^s)+hk\zeta(s)$ where $h$ is the number of ideal classes in $O_K$ and $j_n$ is the number of ideals $I$ in $R$ such that $||I||=n$. The Dirichlet series converges everywhere on the half plane $x>1-(1/[K:\mathbb Q])$ to an analytic function, hence in particular at $s=1$.

Let $h=\rho/k$. I don't understand why $\rho$ is equal to limit $\zeta_K(s)/\zeta(s)$ as $s->1$. Because it seems like the Dirichlet series goes to $0$ and I cannot see why. Hints are welcome.

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    I think it's just because $\zeta(s)$ has a pole at $s = 1$, so $1/\zeta(s) \to 0$ as $s \to 1$. Since $\sum_{n=1}^{\infty} \frac{j_n-hk}{n^s}$ converges to some finite number at $s =1$, then $\frac{1}{\zeta(s)} \sum_{n=1}^{\infty} \frac{j_n-hk}{n^s} \to 0$ as $s \to 1$. The zetas in the second term cancel, leaving you with $hk = \rho$. – 2017-01-19
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    $\zeta_K(s)\overset{def}=\sum_{n=1}^\infty j_nn^{-s} = \sum_{n=1}^{\infty}(j_n-hk)n^{-s}+hk\zeta(s)$ is a tautology, but what you need to prove ([Marcus p.166](http://wstein.org/edu/2010/581b/books/marcus-number_fields.pdf)) is that $\sum_{n < x}(j_n- hk) + \mathcal{O}(x^{1-\frac{1}{[K:\mathbb{Q}]}+\epsilon})$ [and so](https://en.wikipedia.org/wiki/Dirichlet_series#Abscissa_of_convergence) $\sum_{n=1}^{\infty}(j_n-hk)n^{-s}$ is analytic on $Re(s)>1-\frac{1}{[K:\mathbb{Q}]}$ and $\zeta_K(s) \sim hk\zeta(s) \sim \frac{hk}{s-1}$ as $s \to 1$ @SpamIAm – 2017-01-19
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    Thank you both guys. The reasoning of Spam seems correct to me, however I don't understand what user is saying. User, are you saying that the reasoning of spam is wrong? Or just you are explaining the whole setting? – 2017-01-19
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    @Richard When you said "the Dirichlet series converges...", I took that to mean the series $\sum_{n=1}^{\infty} \frac{j_n-hk}{n^s}$. I think user is saying you actually meant $\zeta_K(s)$ converges, in which case you need to do some work to show $\sum_{n=1}^{\infty} \frac{j_n-hk}{n^s}$ converges at $s=1$. – 2017-01-20
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    Well I really meant that the series with $j_n$ converges! Marcus proves it. If you are interested in the topic, please see this question: http://math.stackexchange.com/questions/2105155/two-exercises-on-characters-on-marcus-part-2?noredirect=1#comment4328956_2105155. Nobody is answering and actually I'm starting to think that the exercise could be false, which would be a disaster since it is essential to prove the class number formula. – 2017-01-20

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