# Prove that the sums of the reciprocals of the primes diverge using $\frac{1}{p_{j+1}}+\frac{1}{p_{j+2}}+\frac{1}{p_{j+3}}+\cdots>\frac{1}{2}$

Prove that the sum of the reciprocals of the primes diverge, i.e show that: $$\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}+... Diverges$$

The hint I got:

To show that the sum of the reciprocals of the primes diverges, it’s enough to show that, for any j: $$\frac{1}{p_{j+1}}+\frac{1}{p_{j+2}}+\frac{1}{p_{j+3}}+\cdots>\frac{1}{2}$$

What I have thought:

So basically the hint led me to:

$$\frac{x}{p_{j+1}}+\frac{x}{p_{j+2}}+\frac{x}{p_{j+3}}+\cdots\ge x-N_j(x)$$

• For any number $x$, $N_j(x)$ is the number of positive integers less than or equal to $x$ that have all their prime divisors among the set of the first $j$ primes $\{p_1,p_2,\ldots,p_j\}$.

So, why is $x/p$ greater than or equal to the number of numbers less than or equal to $x$ that are divisible by $p$. What about $x/p + x/q$ where $p$ and $q$ are relatively prime?

And from this how can I prove that the sums of the reciprocals of the primes diverge from using the hints that I gave and the path I took.

• You may want to look at math.stackexchange.com/questions/674725/… also en.wikipedia.org/wiki/… – Gerry Myerson May 16 '15 at 12:06
• @GerryMyerson I'm not sure how that relates to my question. – Cassie May 16 '15 at 12:09
• @GerryMyerson I am actually looking for a specific way to solve this. By $$\frac{1}{p_{j+1}}+\frac{1}{p_{j+2}}+\frac{1}{p_{j+3}}+\cdots>\frac{1}{2}$$ – Cassie May 16 '15 at 12:12
• @Cassie: It concerns a critical step in one argument to show the sum of the reciprocals of primes diverges. – hardmath May 16 '15 at 12:12
• The title of your question, Cassie, is "Prove that the sums of the reciprocals of the primes diverge," and the first sentence is "How can I show that the sum of the reciprocals of the primes diverge?" If what you really want is a proof using that hint, you ought to edit title and body to make this clear. – Gerry Myerson May 16 '15 at 12:18

## 2 Answers

Let's continue along the path you started on from the hint. First, let's address

So, why is $x/p$ greater than or equal to the number of numbers less than or equal to $x$ that are divisible by $p$.

The numbers (positive integers) divisible by $p$ are precisely the numbers of the form $k\cdot p$, where $k \in \mathbb{N}\setminus \{0\}$. Since $p > 0$, such a number is $\leqslant x$ if and only if $k \leqslant \dfrac{x}{p}$, so there are exactly $\biggl\lfloor \dfrac{x}{p}\biggr\rfloor$ positive multiples of $p$ not exceeding $x$, and $\dfrac{x}{p}$ is an upper bound.

What about $x/p+x/q$ where $p$ and $q$ are relatively prime?

That is an upper bound for the count of numbers $\leqslant x$ that are divisible by at least one of $p$ and $q$. There are $\biggl\lfloor \dfrac{x}{p}\biggr\rfloor$ numbers $\leqslant x$ divisibly by $p$, and $\biggl\lfloor\dfrac{x}{q}\biggr\rfloor$ divisible by $q$. But the numbers divisible by both - of which there are $\biggl\lfloor\dfrac{x}{pq}\biggr\rfloor$ - are counted twice, so there are

$$\biggl\lfloor \frac{x}{p}\biggr\rfloor + \biggl\lfloor\frac{x}{q}\biggr\rfloor - \biggl\lfloor \frac{x}{pq}\biggr\rfloor \leqslant \frac{x}{p} + \frac{x}{q}$$

numbers not exceeding $x$ that are divisible by $p$ or by $q$.

Since double-counting (or multiple-counting) can only increase the bound, we then see that the count of numbers not exceeding $x$ that are divisible by some prime $> p_j$, that is, $x - N_j(x)$, is bounded above by

$$\frac{x}{p_{j+1}} + \frac{x}{p_{j+2}} + \frac{x}{p_{j+3}} + \dotsc\,.$$

You have written that inequality down already.

And from this how can I prove that the sums of the reciprocals of the primes diverge.

First finish proving the inequality the hint told you to prove. Dividing the inequality by $x$, you obtain

$$\sum_{k = j+1}^\infty \frac{1}{p_k} \geqslant 1 - \frac{N_j(x)}{x},$$

and that holds for all $x > 0$. So if you find an $x$ such that

$$N_j(x) < \frac{x}{2},$$

you have proved the inequality.

So let us estimate $N_j(x)$. A number that is not divisible by any prime greater than $p_j$ is of the form

$$m = \prod_{k=1}^j p_k^{\alpha_k},$$

where $\alpha_k$ is a non-negative integer for all $k$, and $p_k^{\alpha_k} \leqslant x$. Since $p_k \geqslant 2$, it follows that $2^{\alpha_k} \leqslant x$. Thus you get an upper bound for the exponents in the prime factorisation of all $m \leqslant x$ all of whose prime factors are $\leqslant p_j$, and from the bound on the exponents, you get a bound on $N_j(x)$. From that bound, you can deduce that for all large enough $x$, you have indeed $N_j(x) < \dfrac{x}{2}$. A little bit of calculus knowledge is used there.

Finally, you deduce

$$\Biggl(\bigl(\forall j\bigr)\biggl(\sum_{k=j+1}^\infty \frac{1}{p_k} > \frac{1}{2}\biggr)\Biggr) \implies \sum_{k=1}^\infty \frac{1}{p_k} = +\infty.$$

• What does the \ mean in k∈ℕ∖{0} – Cassie May 16 '15 at 13:59
• Set difference, $A\setminus B = \{ a : (a \in A) \land (a \notin B)\}$. Since we're only looking at positive integers, I've excluded $k = 0$. If you work with the convention that $0\notin \mathbb{N}$, then you can just ignore the "$\setminus \{0\}$" part. – Daniel Fischer May 16 '15 at 14:01
• Also what does this mean: $\biggl\lfloor \dfrac{x}{p}\biggr\rfloor$ – Cassie May 16 '15 at 14:03
• I know that it is the floor function but what does it mean in this case – Cassie May 16 '15 at 14:04
• If you know that it is the floor function, then I don't understand the question. Perhaps "why are we taking the floor"? – Daniel Fischer May 16 '15 at 14:07

Suppose the series converges and converges to $\alpha$, so given $\epsilon >0$ there is a $N \in \mathbb{N}$ such that $\vert \alpha - \sum_{n \leq K} 1/ p_n \vert < \epsilon$. Now let us choose our $\epsilon=1/2$, so we must have $\vert \alpha - \sum_{n \leq K} 1/p_n \vert < 1/2$, but $\alpha - \sum_{n \leq K} 1/p_n = \sum_{n \geq K+1} 1/p_n > 1/2$, so coupling these two inequalities together we get 1/2 < 1/2 a contradiction. So the hint is basically telling us to see that the tail of the series is not converging to zero which means the series diverges. But to prove the hint will be a different matter.

• How do I prove the hint on the path that I took: $$\frac{x}{p_{j+1}}+\frac{x}{p_{j+2}}+\frac{x}{p_{j+3}}+\cdots\ge x-N_j(x)$$ – Cassie May 16 '15 at 13:33