It is well known that $$\sum_{i=1}^{\infty} \frac{1}{p_{i}}$$ diverges, where the $p_{i}$'s are the prime numbers.

Does anybody know a very elementary proof of this result that would be suitable for calculus-level students?

Many thanks.

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    $\begingroup$ See Erdős's proof here. $\endgroup$ Jul 24, 2019 at 4:14
  • $\begingroup$ @RobertIsrael. Wonderfully simple ! BTW. At the age of 18 Erdos found a new proof of Bertrand's "Postulate" (first proven in 1837) $\endgroup$ Jul 24, 2019 at 13:21

3 Answers 3


There is a very nice proof (due to Erdös) that involves a minimum amount of calculus, it is the following:

If $\sum 1/p$ converges then there is an $n$ such that $$1/p_{n+1}+1/p_{n+2} + \dots < \frac{1}{2}$$ chose one $n$ with this property and consider the set $A$ of all integers composed only by the first $n$ primes: $p_1,p_2,\dots,p_n$.

let $A(x)$ be the number of elements of $A$ smaller than $x$, now you can write eny element $a \in A$ as $a=m^2s$ with $s$ squarefree and there are only $2^n$ squarefree integers made by the first $n$ primes. So for all $x$

$$ A(x) \le 2^n\sqrt{x} $$

On the other hand $x-A(x)$ counts the number of integer up to $x$ with at least one prime factor $\ge p_{n+1}$, but this number is at most $$ \left \lfloor\frac{x}{p_{n+1}} \right \rfloor + \left \lfloor\frac{x}{p_{n+2}} \right \rfloor + \dots \le \frac{x}{p_{n+1}} +\frac{x}{p_{n+2}} + \dots < \frac{x}2$$ so we get the two inequalities: $$ A(x) > \frac x2 \quad \text{and} \quad A(x) \le 2^n\sqrt{x} $$ and combining both we get for all $x$: $$ 2^{n+1} > \sqrt{x} $$ but this is clearly false if $x \ge 2^{2n+2}$ leading to a contradiction, so $\sum 1/p$ diverges.

  • $\begingroup$ +1.... According to Wikipedia (see the link in the comment to the Q by @RobertIsrael) this is due to Erdos but there is no citation or reference. $\endgroup$ Jul 24, 2019 at 13:46
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    $\begingroup$ @DanielWainfleet, I hadn't seen the comment, but I have some doubts about my attribution, I had it written in some old notes. I'll try to look later for a reference. $\endgroup$ Jul 24, 2019 at 13:58
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    $\begingroup$ It's in "Proofs from The Book", which refers to "Uber die Reihe $\sum \frac{1}{p}$", Mathematica, Zutphen B 7 (1938), 1-2. $\endgroup$ Jul 24, 2019 at 23:45
  • $\begingroup$ @RobertIsrael Thanks, you are right, I have edited the answer. J. Perott has a proof of Euclid's theorem with a similar flavour: If there are only $n$ primes then there are $2^n$ integers not divisible by a square so for any $N$, $$2^n > N-\sum_p N/p^2 > N(1-(\pi^2/6-1)) > N/3$$ and this is false if $N > 3\cdot 2^n$. $\endgroup$ Jul 25, 2019 at 11:04
  • $\begingroup$ My edit was for typos. In the display following "On the other hand...", the subscripts on the $p$'s were (incorrectly) all $n+1.$ $\endgroup$ May 24, 2021 at 18:11

Define $\zeta(s) = 1+ \frac{1}{2^s}+ \frac{1}{3^s}+ \frac{1}{4^s}+\cdots ,$ where $s \in \mathbb{R}, s> 1$.

We know that $\zeta(s) = \prod\limits_{p \text{ prime}} \left( 1 - \dfrac{1}{p^s} \right)^{-1}.$

Taking $\log$ of both the sides, we have:

$\log \zeta(s) = -\sum\limits_{p \text{ prime}}\log \left( 1 - \dfrac{1}{p^s} \right).$

Expanding $\log$ of right-hand side, we have:

$\log \zeta(s) = \sum\limits_{p \text{ prime}} \left( \dfrac{1}{p^s} \right)+ A(s).$

The term $A(s)$ takes into account the higher powers of primes. Moreover $A(s)$ converges at $s=1$

Taking $\ \displaystyle \lim_{s\to 1}$ , the left hand side is infinity and $A(s)$ is finite, so $\sum\limits_{p \text{ prime}} \left( \dfrac{1}{p} \right)$ diverges.


Note: I improved my answer to show that

$\sum_{p \le n}\frac1{p} \ge \ln(\ln(n))-1 $.

Or Euler's (off the top of my head):

By unique factorization

$\begin{array}\\ \sum_{k=1}^n \dfrac1{k} &\le \dfrac1{\prod_{p \le n}(1-1/p)}\\ &=\prod_{p \le n}(1+\frac1{p-1})\\ &\le\prod_{p \le n}e^{\frac1{p-1}} \qquad\text{since } e^x \ge 1+x\\ &=e^{\sum_{p \le n}\frac1{p-1}}\\ \text{so}\\ \sum_{p \le n}\frac1{p-1} &\ge \ln(\sum_{k=1}^n \dfrac1{k})\\ &\ge \ln(\ln(n)) \qquad\text{by integral test}\\ \end{array} $

Since $\dfrac1{p-1} \le \dfrac{2}{p}$, we have $\sum_{p \le n}\frac1{p} \gt \frac12\ln(\ln(x)) $.

Here is an improvement on this last paragraph.

$\begin{array}\\ \sum_{p \le n}\frac1{p-1} -\sum_{p \le n}\frac1{p} &=\sum_{p \le n}(\frac1{p-1}-\frac1{p})\\ &=\sum_{p \le n}\frac1{p(p-1)}\\ &\le\sum_{k=2}^{n}\frac1{k(k-1)}\\ &\lt 1\\ \end{array} $


$\sum_{p \le n}\frac1{p} \gt \sum_{p \le n}\frac1{p-1}-1 \ge \ln(\ln(n))-1 $.

  • $\begingroup$ We can simplify this somewhat: (1). Each term $1/k$, for $1\le k\le n,$ occurs at least once in the complete expansion of $F(n)=\prod_{p\le n}\sum_{j=0}^{n}p^{-j}. $(Exactly once by unique factorization,but we don't need it). So $\sum_{k=1}^n(1/k)\le F(n)<\prod_{p\le n}(1-1/p)^{-1}.$ So $\prod_{p\le n}(1-1/p)\to 0.$... So if $p_j$ is the $ j$th prime then $\prod_{j\in \Bbb N}(1-1/p_j)=0.$....A basic result on series is that if $a_j<1$ for all $j\in \Bbb N$ then $\sum_{j\in \Bbb N} a_j$ converges iff $\prod_{j\in \Bbb N}(1-a_j)>0.$ $\endgroup$ Jul 24, 2019 at 13:01
  • $\begingroup$ A basic "companion" result is that if $a_j\ge 0$ for all $j$ then $\sum_{j\in \Bbb N}a_j<\infty$ iff $\prod_{j\in \Bbb N}(1+a_j)<\infty.$ Both of these basic results can be briefly shown, even without calculus or logarithms (and more easily with'em). $\endgroup$ Jul 24, 2019 at 13:10
  • $\begingroup$ In my first comment it should say $0\le a_j<1$ instead of $a_j<1.$ $\endgroup$ Jul 24, 2019 at 13:27
  • $\begingroup$ I wanted to get the rate of divergence, not just show that it diverges. I found it interesting that $\sum \frac1{p-1}$ occurred and felt good that I was able (in the addition at the end) to convert this to $\sum \frac1{p}$. $\endgroup$ Jul 24, 2019 at 15:46
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    $\begingroup$ Like most of my answers, I did this off the top of me head without looking anything up, so there certainly is a chance I made a mistake. However, I just looked at it again and don't see anything wrong. $\endgroup$ Jul 24, 2019 at 22:54

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