Merten's Theorem for Arithmetic Progressions
This content was prepared as part of a course project for Analytic Number Theory, conducted under the guidance of Prof. Biswajyoti Saha. The primary objective is to analyze equation (2.6) in \cite{williams1974merten}. Although we found the reference to equation (2.6), we were unable to fully understand it. As a result, we did some further investigation and provided a clarification of the equation.
Lemma
(Tao, 2014)
Let $ \chi $ be a non-principal character of modulus $ k $, and let $ f $ be an arithmetic function that is monotone on an interval $ [x,y] $. Then
Proof
We begin by observing that we can assume $ f $ is monotone non-increasing, as the argument relies only on the behavior of $ f $, not its specific ordering. Next, we modify $ x $ and $ y $ by rounding $ x $ up and $ y $ down to the nearest multiples of $ k $. This ensures $ x $ and $ y $ are multiples of $ k $, which simplifies the analysis. Consequently, the sum on the left-hand side can now be expressed as a sum over blocks of size $ k $:
\[\sum_{x\leq n<y} f(n)\chi(n) = \sum_{j} \sum_{jk \leq n < (j+1)k} f(n)\chi(n),\]where $ j $ ranges over integers satisfying $ x\leq jk < (j+1)k <y $.
Now, using the fact that $ \chi $ is non-principal, we know it is orthogonal to the principal character. This implies:
\[\sum_{jk \leq n < (j+1)k} \chi(n) = 0.\]Therefore, we can rewrite each inner sum as:
\[\sum_{jk \leq n < (j+1)k} f(n)\chi(n) = \sum_{jk \leq n < (j+1)k} (f(n) - f(jk))\chi(n).\]| Next, we use the trivial bound $ | \chi(n) | \leq 1 $ to estimate the magnitude of this sum. Since $ f $ is monotone non-increasing, |
Thus, the entire sum can be bounded as:
\[\left| \sum_{jk \leq n < (j+1)k} f(n)\chi(n) \right| \leq k \big(f(jk) - f((j+1)k)\big).\]Finally, summing over $ j $, we observe that the terms $ f(jk) - f((j+1)k) $ form a telescoping series. This causes most terms to cancel, leaving a bound dependent on the endpoints. The claim follows directly from this telescoping argument.
Corollary
Let $ \chi $ be a non-principal Dirichlet character of modulus $ k $. Then
\[\sum_{n\leq x} \frac{\Lambda(n)\chi(n)}{n} = \mathcal{O}_{\chi}(1),\]for any $ x\geq 1 $.
Proof
From the lemma, we have that
\[\sum_{n\leq x} f(n)\chi(n) = \mathcal{O}_{k}(|f(x)|).\]Therefore,
\[\sum_{n\leq x} \frac{\Lambda(n)\chi(n)}{n} = \mathcal{O}_k(\Lambda(x)/x) = \mathcal{O}_k(1).\]Lemma
Let $ \chi $ be a non-principal Dirichlet character of modulus $ k $. Then
\[\sum_{p\geq x} \frac{\chi(p)}{p} = \mathcal{O}\left(\frac{1}{\log{x}}\right).\]Proof
We start by rewriting the summation on the left-hand side. Using the Abel Summation Formula, we express:
\[\sum_{x \leq p \leq y} \frac{\chi(p)}{p} = \frac{1}{\log y} \sum_{x \leq p \leq y} \frac{\chi(p)\log p}{p} + \int_x^y \frac{1}{t \log^2 t} \sum_{x \leq p \leq t} \frac{\chi(p)\log p}{p} \, dt.\]To proceed, we need an asymptotic estimate for
\[\sum_{p \leq x} \frac{\chi(p)\log p}{p}.\]By combining this result with the previous corollary, it follows that:
\[\sum_{p \leq x} \frac{\chi(p)\log p}{p} = \mathcal{O}_{\chi}(1).\]Hence, for all $ x \leq t \leq y $, we have:
\[\sum_{x \leq p \leq t} \frac{\chi(p)\log p}{p} = \mathcal{O}(1).\]Substituting this back, we find:
\[\sum_{x \leq p \leq y} \frac{\chi(p)}{p} = \mathcal{O}\left(\frac{1}{\log y} + \frac{1}{\log x}\right).\]Finally, letting $ y \to \infty $ gives the desired result.
Theorem
Let $ \chi $ be a non-principal Dirichlet character of modulus $ k $. Then
\[\prod_{p\leq x} \left(1-\frac{\chi(p)}{p}\right) = \frac{1}{L(1,\chi)} + \mathcal{O}\left(\frac{1}{\log{x}}\right),\]as $ x\to \infty $.
Proof
Since
\[L(1,\chi) = \prod_{p}\left(1-\frac{\chi(p)}{p}\right)^{-1},\]we note that:
\[L(1,\chi)\prod_{p\leq x}\left(1-\frac{\chi(p)}{p}\right) = \prod_{p>x} \left(1-\frac{\chi(p)}{p}\right)^{-1}.\]It suffices to show that:
\[\log\left(L(1,\chi)\prod_{p>x} \left(1-\frac{\chi(p)}{p}\right)^{-1}\right) = \mathcal{O}\left(\frac{1}{\log{x}}\right).\]Now,
\[\log{\prod_{p>x} \left(1-\frac{\chi(p)}{p}\right)^{-1}} = \sum_{p>x}\frac{\chi(p)}{p} + \sum_{p>x}\sum_{m\geq 2}\frac{\chi^m(p)}{mp^m}.\]Using the previous lemma, we obtain:
\[\sum_{p>x} \frac{\chi(p)}{p} = \mathcal{O}\left(\frac{1}{\log{x}}\right).\]Similarly,
\[\sum_{p>x}\sum_{m\geq 2}\frac{\chi^m(p)}{mp^m} = \mathcal{O}(1/x).\]Substituting these results, we conclude:
\[\log{\prod_{p>x} \left(1-\frac{\chi(p)}{p}\right)^{-1}} = \mathcal{O}\left(\frac{1}{\log{x}}\right),\]completing the proof.