Is this series known to converge, and if so, what does it converge to (if known)?

Where $p_n$ is prime number $n$, and $p_1 = 2$,

$$\sum\limits_{n=1}^\infty \frac{1}{p_n}$$

  • 9
    $\begingroup$ A cultural note: Euler was (as far as I know) the first person to observe that this series diverges (and without assuming a priori that it was an infinite series), thus obtaining a new proof of the infinitude of the primes. This was a precursor to Dirichlet's work in which he proved the infinitude of primes in arithmetic progressions, and then Riemann, Hadamard, and de la Vallee Poussin's work leading to the prime number theorem. $\endgroup$
    – Matt E
    Dec 31, 2010 at 4:28
  • 6
    $\begingroup$ Unfortunately, the same technique does not apply for other problems. For instance, the sum of reciprocals of the twin primes converges and the existence of infinitely many twin primes remains open. $\endgroup$
    – lhf
    Dec 31, 2010 at 6:57
  • $\begingroup$ For fun try summing $\sum_2^\infty P(x)$ where P(x) is the Prime Zeta function. The actual sum of reciprocal primes is similar to the harmonic numbers which diverge. $\endgroup$ Sep 25, 2021 at 2:44
  • $\begingroup$ @lhf the sum of reciprocals of twin primes is almost certainly the wrong sum to use. The Bateman-Horn conjecture on simultaneous prime values of polynomials $f_1(x),\ldots,f_r(x)$ implies that, when the $f_i(x)$ fit the hypotheses of that conjecture, $\sum_{n \leq x, n \in S} (\log n)^{r-1}/n \sim ({\rm const.})\log\log x$, where $S$ is the set of $n$ s.t. $f_1(n),\ldots,f_r(n)$ are all prime. For $r = 1$ we get $\sum_{p \leq x} 1/p \sim ({\rm const.})\log\log x$, and for $r=2$ we get $\sum_{p \leq x, p \in T} (\log p)/p \sim ({\rm const.})\log\log x$ where $T$ is the set of twin primes. $\endgroup$
    – KCd
    Jun 23, 2022 at 16:55

3 Answers 3


No, it does not converge. See this: Proof of divergence of sum of reciprocals of primes.

In fact it is known that $$\sum_{p \le x} \frac{1}{p} = \log \log x + A + \mathcal{O}(\frac{1}{\log^2 x})$$

Related: Proving $\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} \geq \frac{1}{2}\log{x} -\log{\log{x}}$


I would like to note that this implies that according the Müntz-Szász Theorem that every continuous function in $[0,1]$ is a uniform limit of polynomials whose exponents are prime numbers!

  • 4
    $\begingroup$ Could that implication possibly have any applications? $\endgroup$
    – KCd
    Dec 31, 2010 at 22:09
  • $\begingroup$ @KCd: I don't know, I'm more into analysis and there (at least the part I do) prime numbers are not that important, maybe it can be useful in analytic number theory. $\endgroup$
    – JT_NL
    Jan 1, 2011 at 15:13

Let's start with three lemmas:

  1. Suppose $A\subseteq\{1,2,3,\ldots\}$ and $\sum\limits_{n\in A} \dfrac 1 n < \infty$. Then $\sum\limits_{n\in B} \dfrac 1 n <\infty$ where $B$ is the closure of $A$ under multiplication.

  2. The closure of the set of primes under multiplication is all of $\{1,2,3,\ldots\}$.

  3. $\sum\limits_{n=1}^\infty \dfrac 1 n = \infty$.

The second lemma is obvious. The third has a number of well known simple proofs. Here is one of those: \begin{align} & \frac 1 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \frac 1 6 + \cdots \tag 1 \\[10pt] = {} &\left(\frac 1 1 + \frac 1 2\right) + \left(\frac 1 3 + \frac 1 4\right) + \left(\frac 1 5 + \frac 1 6\right) + \cdots \\[10pt] \ge {} & \left(\frac 1 2 + \frac 1 2 \right) + \left( \frac 1 4 + \frac 1 4 \right) + \left( \frac 1 6 + \frac 1 6 \right) + \cdots \tag 2 \\[10pt] = {} & \frac 1 1 + \frac 1 2 + \frac 1 3 + \cdots \end{align} The inequality on line $(2)$ is strict if the sum on line $(1)$ is finite, and that leads us to a contradiction. ${}\qquad\blacksquare$

The proof of lemma 1 is most of the work; here it is:

\begin{align} & \sum_{n\in B} \frac 1 n \le \overbrace{\sum_{\begin{smallmatrix} C\subseteq A \\[2pt] C \text{ is finite} \end{smallmatrix}} \prod_{k\in C} \frac 1 k = \prod_{a\in A} \sum_{x=0}^\infty \frac 1 {a^x}}^\text{factoring -- see below} = \prod_{a\in A} \frac 1 {1-\frac 1 a} \\[10pt] = {} & \exp \sum_{a\in A} - \log\left( 1 - \frac 1 a\right) \le \exp \sum_{a\in A} \frac 1 a < \infty. \end{align} (As "Pipicito" points out in a comment below, some members of the set $B$ may occur more than once in the sum below and that is why $\text{“}{\le}\text{''}$ rather than $\text{“}{=}\text{''}$ should appear in the first step above.)

Here's the factorization in more detail: Let $A=\{a_1,a_2,a_3,\ldots\}$. Then the product to the right of $\text{“}{=}\text{''}$ under the $\overbrace{\text{overbrace}}$ above is \begin{align} & \left( 1 + \frac 1 {a_1} + \frac 1 {a_1^2} + \frac 1 {a_1^3} + \cdots \right) \\ \times {} & \left( 1 + \frac 1 {a_2} + \frac 1 {a_2^2} + \frac 1 {a_2^3} + \cdots \right) \\ \times {} & \left( 1 + \frac 1 {a_3} + \frac 1 {a_3^2} + \frac 1 {a_3^3} + \cdots \right) \\ \times {} & \quad \cdots \cdots \\ \vdots~ \end{align} When you expand the product, you multiply a term from the first factor, a term from the second factor, a term from the third factor, etc., but all except finitely many of those are $1$. The reason all but finitely many are $1$ is that if you multiply infinitely many non-$1$s, then the product is $0$, since its a product of infinitely many positive numbers less than $1/2$. Then you add up all possible such finite products, and that gives you the sum to the left of $\text{“}=\text{''}$ under the $\overbrace{\text{overbrace}}$ above.

  • 2
    $\begingroup$ In the proof of the first lemma, the first equality in $\sum_{n\in B} \frac 1 n = \cdots \;$ should be a $\leq$ because the elements in $B$ may not be uniquely expressed as products of elements in $A$. For example, if $A = \{ 1, 2, 4\}$ then any element in $B$ greater than or equal to $4$ admits at least two essentially different writings as products of elements in $A$. $\endgroup$
    – Pipicito
    May 11, 2016 at 20:08
  • 2
    $\begingroup$ Also, the proof should be refined. If you admit only $C$ to be a finite subset of $A$ you are missing elements in the closure of $A$ under multiplication. You should admit multisets to permit repetitions. But in that case, the $1$ should be excluded when picking the finite multisubsets of $A$ because, if not, then you could pick $\{ 1\}$, $\;\{ 1, 1\}$, $\;\{ 1,1,1\}$, $\; \dots \;$ but that will always make the bounding sum divergent. $\endgroup$
    – Pipicito
    May 11, 2016 at 20:08

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