# Proof of Dirichlet L-function Euler Product formula (from Fourier Analysis by Stein)

On page 260 of Stein and Shakarchi's "Fourier Analysis," there's a proof of the Dirichlet product formula:

$\sum_{n}\frac{\chi(n)}{n^s}=\Pi_{p}\frac{1}{1-\chi(p)p^{-s}}$

where $s>1$, $\chi$ is a Dirichlet character of $\mathbb{Z}^{*}(q)$ ($q$ a positive integer), and the product is over all primes.

The proof goes as follows:

We know the LHS of the formula converges. Say $L=\sum_{n}\frac{\chi(n)}{n^s}$.

We also know by some arguments that I will not write out here that the RHS of the formula also converges, say $\Pi=\Pi_{p}\frac{1}{1-\chi(p)p^{-s}}$.

Now define $S_N=\sum_{n\leq N} \chi(n)n^{-s}$ and $\Pi_N=\Pi_{p\leq N}\frac{1}{1-\chi(p)p^{-s}}$.

Also define $\Pi_{N,M}=\Pi_{p\leq N}(1+\frac{\chi(p)}{p^s}+\cdots+\frac{\chi(p^M)}{p^{Ms}})$

Now fix $\epsilon>0$ and we can choose $N$ large enough that $|S_N-L|<\epsilon$ and $|\Pi_N-\Pi|<\epsilon$. These are mere consequences of the fact that $S_N\rightarrow L$ and $\Pi_N\rightarrow \Pi$.

Now my question arises in the next part. We've fixed an $\epsilon$ and we've chosen an $N$. The proof proceeds by saying that we can select $M$ large enough so that

1) $|S_N-\Pi_{N,M}|<\epsilon$ and

2) $|\Pi_{N,M}-\Pi_N|<\epsilon$.

The second inequality is clear since the factors of the product $\Pi_{N,M}$ each converge to $\frac{1}{1-\chi(p)p^{-s}}$ if we let $M\rightarrow\infty$.

The first inequality on the other hand doesn't seem so clear. The book says that it follows from using the fundamental theorem of arithmetic and the fact that the Dirichlet characters are multiplicative (i.e. $\chi(a)\chi(b)=\chi(ab)$).

I can see what kind of argument is required: Since the product $\Pi_{N,M}$ is finite, one could easily imagine what the product would look like when fully multiplied out into a sum. There would be very many terms in the sum, and for $M$ large enough, the sum should include every single term of $S_N$...and many more! My concern is that if we just let $M$ be huge (say..much bigger than our chosen $N$), we'd have so many more extra terms in the expanded version of $\Pi_{N,M}$ than just the terms of $S_N$. Basically I'm trying to find an explanation as to why a large $M$ can still guarantee $|S_N-\Pi_{N,M}|<\epsilon$ given all those extra terms.

Any explanations would be appreciated! I am fully aware that there are many alternative proofs (I particularly fancy the one here http://www.maths.bris.ac.uk/~madjmdc/Intro%20to%20L-functions%20-%20Steuding.pdf). However, I'd simply like to understand the logic behind the step that I'm confused about in the proof above.

If you're curious, that's basically the end of the proof since (assuming we have everything from above) we have

$|L-\Pi|\leq |L-S_N|+|S_N-\Pi_{N,M}|+|\Pi_{N,M}-\Pi_{N}|+|\Pi_N-\Pi|<4\epsilon$.

• by expanding the product, using the unique factorization theorem $\prod_{p} (1-\chi(p)p^{-s})^{-1}$ and $\sum_{n=1}^\infty \chi(n) n^{-s}$ are the same formal series, and $\{\chi(n) n^{-s}\}_{n \in \mathbb{N}}$ is absolutely summable for $Re(s) > 1$, hence the order of summation doesn't matter, and $F(s) = \prod_{p} (1-\chi(p)p^{-s})^{-1} = \sum_{n=1}^\infty \chi(n) n^{-s}$ – reuns May 24 '16 at 14:58
• alternatively for $Re(s) > 1$ you can write $\sum_{n=1}^\infty \chi(n) n^{-s}- \prod_{p \le k} (1-\chi(p)p^{-s})^{-1} = \sum_{n \in \mathbb{N}_k} \chi(n) n^{-s}$ where $\mathbb{N}_k$ is the set of integers having at least one prime factor $> k$. proving that $\sum_{n \in \mathbb{N}_k} \chi(n) n^{-s} \to 0$ as $k \to \infty$ is enough for proving that $$\sum_{n=1}^\infty \chi(n) n^{-s} = \lim_{k \to \infty} \prod_{p \le k} (1-\chi(p)p^{-s})^{-1} = \prod_{p } (1-\chi(p)p^{-s})^{-1}$$ – reuns May 24 '16 at 15:01
• Thank you for your insight! However, I am merely looking for an explanation for the logic laid out by the book. I'll make an edit to make that clear :) – Fozz May 24 '16 at 15:17
• what I mean is : if you understand what I wrote, then you can easily answer yourself to your question, if you don't, tell me what you have a problem with (and the proof uses that $|\chi(n)| \le 1$ for the absolute convergence when $Re(s) > 1$) – reuns May 24 '16 at 15:20

If we choose $M$ sufficiently large, then for each term in $S_N$, there exists a corresponding term in $\Pi_{N,M}$. Hence we have $$\vert\Pi_{N,M}-S_N\vert=\vert\sum_{n\in A}\frac{\chi(n)}{n^s}\vert$$ where $A$ is a subset of $\{N+1,N+2,\cdots, \infty\}$.

Note that If the textbook choose $N$ (to make sure $\vert S_N-S\vert<\epsilon)$ in the way where $\sum_{n=N+1}^{\infty}\frac{1}{n^s}\le \epsilon,$ then $\vert\sum_{n\in A}\frac{\chi(n)}{n^s}\vert\le\sum_{n=N+1}^{\infty}\frac{1}{n^s}<\epsilon$, and this what we desire.

By the way, I am also learning this book recently and stuck in some exercises. I find the biggest problem is that there is no solution for all of the exercises. Do you have a solution? If so, I would appreciate it very much if you would offer one copy to me !!!

• The exercises in this book – S.Wei Nov 16 '16 at 8:19
• what exercice in particular ? – reuns Nov 16 '16 at 8:30
• I am stuck in the Exercise 14 of Chapter 8, and I do not know how to prove the identity in this exercise. – S.Wei Nov 16 '16 at 8:44
• $\sum_{n=1}^\infty \frac{e^{inx}}{n} = \lim_{r \to 1^-}\sum_{n=1}^\infty r^n\frac{e^{inx}}{n} = \lim_{r\to 1^-} -\log(1-re^{ix})$ – reuns Nov 16 '16 at 9:10
• Could provide more detail about the proof? – S.Wei Nov 16 '16 at 14:05

I for one feel that the problem would make more sense if we consider the sum $\sum_{n \in A_{N}} \frac{\chi(n)}{n^{s}},$ where $A_{N} \equiv \{k : p|k \implies p \leq N \}$, instead of the $\sum_{n \leq N}$.

Note that $\sum_{n \in A_{N}} \frac{\chi(n)}{n^{s}} = \prod_{p \leq N}(1 - \chi(p)p^{-s})^{-1}.$

My answer (and any other answers above) seems to use $s > 1$ or absolute convergence. For mine, s has to be greater than 1 so that $\lim_{N \to \infty}\sum_{n \in A_{N}} \frac{\chi(n)}{n^{s}} = \lim_{N \to \infty}\sum_{n \leq N} \frac{\chi(n)}{n^{s}}.$