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?

  • $\begingroup$ @JoshuaStucky but how is the error term dependent upon $x$? $\endgroup$
    – Anish Ray
    Mar 12 at 23:29
  • $\begingroup$ 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. $\endgroup$ Mar 12 at 23:32
  • $\begingroup$ Sorry, I didn't actually finish my comment. $\endgroup$ Mar 12 at 23:33
  • $\begingroup$ @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. $\endgroup$
    – Anish Ray
    Mar 12 at 23:36
  • $\begingroup$ 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?) $\endgroup$
    – tomos
    Mar 13 at 14:10

1 Answer 1


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 $t<q$. 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.


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