# Known upper bounds for $\sum_\limits{p\equiv a \pmod q, p\leq x}\frac{1}{p}$

I am trying to deduce an upper bound for $$\sum_\limits{p\equiv a \pmod q, p\leq x}\frac{1}{p}$$ where $$p$$ is a prime and $$(a,q)=1$$. Using Theorem 1 of Carl Pomerance's article on amicable numbers (which uses simple partial summation), one straightaway obtains, for $$x\ge3$$ and an absolute constant $$A>0$$, $$\sum_\limits{p\equiv a \pmod q, p\leq x}\frac{1}{p}<\frac{3}{2}+\frac{\log\log x}{\phi(q)}+\frac{\log q}{\phi(q)}+\frac{A+1}{q\phi(q)}+\frac{A}{q^2\phi(q)}.$$ But this error term doesn't suit my needs since I want the minor terms to be in the form $$E(x,q)$$. So, one can probably derive a better estimate using the Large-Sieve inequality as stated in Theorem 4.2 of Montgomery's book on multiplicative number theory but this gives an even worse estimate of the form $$O\left(\frac{\sqrt{x}}{\sqrt{\log x}}\right)$$ since I guess, the sum $$\sum\limits_{n}|a_n|^2$$ is convergent. But the large sieve inequality is supposed to give a better or maybe an optimal bound of the form $$\frac{2\log\log x}{\phi(q)}+E(x,q)$$ if we choose suitable (smooth) weight functions to get small Fourier coefficients.

But I am unable to see what could be those weight functions and how to modify the large sieve inequality accordingly. On the other hand, since this seems to be such a well-known mathematical expression, does there exist any previous work that deals with this using the large sieve inequality? If yes, then could somebody refer it to me?

• @JoshuaStucky but how is the error term dependent upon $x$? Mar 12 at 23:29
• What do you mean "I want the error term to be in the form $E(x,q)$?" In the paper of Pomerance that you reference, he proves (Theorem 1) that $$\bigg| \sum_{\substack{p\leq x\\ p\equiv a \text{(mod q)} }} \frac{1}{p} - \frac{\log\log x}{\phi(q)} \bigg| \leq A$$ where $A$ is an absolute constant independent of $a$ and $q$. One certainly cannot replace $A$ by a function tending to 0 with $x$, since there is genuinely a constant term in the asymptotic expansion of the sum. Mar 12 at 23:32
• Sorry, I didn't actually finish my comment. Mar 12 at 23:33
• @JoshuaStucky I see. Then, maybe I should reformulate my question. I need the minor terms of the estimate to be dependent on $x$ as well. Mar 12 at 23:36
• i think it's possible that you're confused as to what exactly you want: you ask for an upper bound but the Pomerance paper and Joshua Stucky's comment tells you more than an upper bound - they give you an actual asymptotic. (i could be wrong, but in that case can you give more information as to what kind of result you expect?) Mar 13 at 14:10

The OP was very confusing for me, but I think this might be helpful.

Let $$a,q\in\mathbb{Z}$$, $$x\geq 3$$, and $$A>0$$. Suppose that $$q\geq 1$$, $$\gcd(a,q)=1$$, and $$q\leq (\log x)^A$$. There exist constants $$c_{a,q}>0$$ (effectively computable) and $$c_A>0$$ (effectively computable when $$A<1$$) such that

$$\sum_{\substack{p\leq x \\ p\equiv a\pmod{q}}}\frac{1}{p}=\frac{\log\log x}{\varphi(q)}+c_{a,q}+O(e^{-c_A\sqrt{\log x}}).$$

The implied constant is effectively computable when $$A<1$$. This follows from the classical zero-free regions for Dirichlet $$L$$-functions along with some other delicate estimates.

It seems to me that what the OP is interested in the sharp form of the Brun-Titchmarsh theorem proved by Montgomery and Vaughan: If $$x>0$$, $$a,q\in\mathbb{Z}$$, $$\gcd(a,q)=1$$, and $$x>q$$, then

$$\pi(x;q,a):=\sum_{\substack{p\leq x \\ p\equiv a\pmod{q}}}1\leq \frac{2x}{\varphi(q)\log(2x/q)}.$$

Montgomery and Vaughan do use the analytic large sieve inequality (in contrast with other large sieve inequalities). Using the MV bound along with partial summation, we have

$$\sum_{\substack{p\leq x \\ p\equiv a\pmod{q}}}\frac{1}{p}=\frac{\pi(x;q,a)}{x}+\int_1^x \frac{\pi(t;q,a)}{t^2}dt\leq \frac{2}{\varphi(q)}\log\Big(\frac{\log(\frac{2x}{q})}{\log 2}\Big)+\frac{2}{\varphi(q)\log(\frac{2x}{q})}+\int_1^q \frac{\pi(t;q,a)}{t^2}dt.$$

Note that we cannot use MV to estimate $$\pi(t;q,a)$$ once $$t. We only know how to estimate $$\pi(t;q,a)$$ trivially in this range: $$\pi(t;q,a) < \pi(t)$$ (the total number of primes up to $$t$$). Using MV with $$q=1$$, we have

$$\int_1^q \frac{\pi(t;q,a)}{t^2}dt\leq \int_{3/2}^q \frac{2t}{t^2\log(2t)}dt=2\log\Big(\frac{\log(2q)}{\log 3}\Big).$$

However, Pomerance's result is far better than what you get "from the large sieve inequality" if $$x$$ is small relative to $$q$$, and the result using zero-free regions is far better than what you get "from the large sieve inequality" if $$x$$ is large relative to $$q$$. Either way, "the large sieve inequality" seems to be the wrong way to approach the sum of the reciprocals of the primes; it's too inefficient.