Convergence Of Infinite Products

Infinite Series (Sums)

At school we learned about infinite series.

$$  \sum_{n=1}^\infty a_n = a_1+ a_2 + a_3 + \ldots  $$

If the partial sum $\sum_{n=1}^{x} a_n$ tends to a limit as $x \rightarrow \infty$, we can say the series converges to a finite sum. There are many tests that can be used to see if a series converges or not, such as the popular ratio test:

$$ \lim_{n \rightarrow \infty} | \frac{a_{n+1}}{a_n} | <1 $$

If we think about the series $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots$, the ratio test gives us $|\frac{ 2^{n-1} }{ 2^{n} }| = \frac{1}{2}$, confirming the series is absolutely convergent.

Infinite Products

Infinite products don't seem to be discussed as much as series.

$$ \prod_{n=1}^\infty a_n = a_1 \times a_2 \times a_3 \times \ldots  $$

When do such infinite products converge? Let's think about this before jumping to the answer.

Thoughts

It seems intuitive that if the elements $a_n$ are getting larger, the product grows and diverges. The following example gets ever larger and doesn't converge. 

$$  2 \times 3 \times 4 \times 5 \times \ldots $$

Let's look at another example.

$$ 0.5 \times 0.55 \times 0.555 \times 0.5555 \times \ldots $$

Here the elements are getting larger, but are individually less than 1, so it isn't immediately obvious whether the product converges or not.

What if the product contains a zero? That sounds easy. A single zero would immediately make the overall product zero. 

But what about this product? 

$$ -2 \times -1 \times 0 \times 1 \times 2 \times \ldots $$

With the zero, we'd suggest the product is zero overall. But without that zero, the product gets larger and diverges. Is the single zero somehow 'powerful' enough to cancel the infinity?

Definition

A consensus has emerged around infinite products which may, or may not, seem natural.

If the partial product $\prod_{n=1}^{x} a_n$ converges to a non-zero limit $P$ as $x \rightarrow \infty$, we say the infinite product converges to $P$.

This makes sense but the fact that the limit needs to be non-zero sets infinite products apart from infinite series.

If a finite number of zeros $a_n$ can be removed from a product, and the remaining product converges, we say the infinite product converges to 0.

This is rather specific, and means that we can't say a product with infinitely many zero elements converges. For example, we can't say the following converges:

$$  1 \times 0 \times 1 \times 0 \times 1 \times 0 \times 1 \times \ldots $$

Let's look at some examples to bring these ideas to life.

Example 1

Does the following infinite product converge or diverge?

$$ \prod_{n=1}^{\infty}(1+\frac{1}{n}) $$

If we consider the partial sum, and look at at the first few elements of this product, we'll see a pattern.

$$ \begin{align} \prod_{n=1}^{N}(1+\frac{1}{n}) &=  \prod_{n=1}^{N}(\frac{n+1}{n}) \\ \\ &= \frac{\cancel{2}}{1} \times \frac{\cancel{3}}{\cancel{2}} \times \frac{4}{\cancel{3}} \times  \ldots \times \frac{N+1}{N} \\ \\ &= N + 1 \end{align} $$

As $N \rightarrow \infty$, this partial sum tends to $\infty$, so the infinite product diverges. This isn't surprising when we look at each element and that they are all above 1.

Example 2

Let's now look at an example where each element is less than 1. Notice $n$ starts at 2, not 1. 

$$ \prod_{n=2}^{\infty}(1-\frac{1}{n}) $$

Again, if we consider the partial sum, and look at at the first few elements of this product, we'll see a pattern.

$$ \begin{align} \prod_{n=2}^{N}(1-\frac{1}{n}) &=  \prod_{n=2}^{N}(\frac{n-1}{n}) \\ \\ &= \frac{1}{\cancel{2}} \times \frac{\cancel{2}}{\cancel{3}} \times \frac{\cancel{3}}{4} \times  \ldots \times \frac{N-1}{N} \\ \\ &= \frac{1}{N} \end{align} $$

As $N \rightarrow \infty$, this partial sum tends to 0. Because this limit is not a non-zero limit, as required by the definitions above, we say the infinite product diverges. 

Turning Products Into Sums

We can convert a product into a sum using logarithms.

$$ \begin{align} \ln (\prod_{n=1}^\infty a_n) &= \ln( a_1 \times a_2 \times a_3 \times \ldots)  \\ \\ &= \sum_{n=1}^\infty \ln(a_n) \end{align}$$

We can say the infinite product converges if the sum $\sum_{n=1}^\infty \ln(a_n)$ converges. 

If that product tends to zero, this sum tends to $-\infty$, which is why we say the infinite product diverges to zero, not converges to zero.

A Different Test For Convergence

If we can rewrite the product elements as $(1 + a_n)$ where $a_n >0$ we can use a simpler convergence test.

$$ \begin{align} \prod_{n=1}^\infty (1 + a_n) &\leq  \prod_{n=1}^\infty e^{ a_n} \\ \\ &=   e^{\sum_{n=1}^\infty a_n} \end{align} $$

The inequality comes from $e^x=1 + x + O(x^2)$. We can make the following conclusion.

If $\sum_{n=1}^\infty a_n$ converges, then so does the product $\prod_{n=1}^\infty (1 + a_n)$, as long as $a_n >0$.

This test allows us to easily say the following infinite series converges:

$$ \prod_{n=1}^\infty (1 + \frac{1}{n^2}) $$

We can say this because we know $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges.

Infinite Products Over Complex Numbers

There is an equivalent form of the previous test for complex $z_n$.

If $\sum_{n=1}^\infty |a_n|$ converges, then so does the product $\prod_{n=1}^\infty (1 + a_n)$, as long as $a_n \neq -1$.

Mathologer Video on Euler's Product Formula

The Mathologer youtube channel is a treasure trove of mathematics made fun and accessible. His video on the Euler product formula is particularly clear and worth watching.

He even includes some really nice and easily derived consequences of Euler's product formula, including:

• there are infinitely many primes because the harmonic series diverges

• there are infinitely many primes because $\pi$ is not a rational number

• probability that two randomly chosen integers are coprime is $\frac{6}{\pi^2}$

Euler's Golden Bridge

Euler was the first to find a connection between the world of primes and the world of ordinary counting numbers. Many insights about the primes have been revealed by travelling over this `golden bridge'.

Let's recreate Euler's discovery for ourselves. 

Video: [youtube]

Slides: [pdf]

Reimann Zeta Function

We know the harmonic series $\sum1/n$ diverges. We also know the series $\sum1/n^{2}$ converges. 

It's natural to ask for which values of $s$ the more general series, known as the Riemann Zeta function $\zeta(s)$, converges.

$$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^{s}}=\frac{1}{1^{s}}+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\ldots$$

Series like $\sum1/n^{3}$ and $\sum1/n^{4}$ converge because each term is smaller than the corresponding one in $\sum1/n^{2}$. Less obvious is when $s<2$.

A separate blog post presents a short proof that $\zeta(s)$ converges for $s>1$. 

Sieving The Zeta Function

Let's write out the zeta function again, noting that $1^{s}=1$.

$$\zeta(s)=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\frac{1}{5^{s}}+\frac{1}{6^{s}}+\ldots$$

We can divide this series by $2^{s}$. 

$$\frac{1}{2^{s}}\zeta(s)=\frac{1}{2^{s}}+\frac{1}{4^{s}}+\frac{1}{6^{s}}+\frac{1}{8^{s}}+\frac{1}{10^{s}}+\frac{1}{12^{s}}\ldots$$

These denominators are multiples of $2^{s}$. By subtracting these terms from $\zeta(s)$, we sieve out terms with these multiples of $2^{s}$. 

$$(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{3^{s}}+\frac{1}{5^{s}}+\frac{1}{7^{s}}+\frac{1}{9^{s}}+\frac{1}{11^{s}}+\ldots$$

This dividing and subtracting of infinite series is only valid because they are absolutely convergent for $s>1$.

Let's now divide this series by $3^{s}$.

$$\frac{1}{3^{s}}\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=\frac{1}{3^{s}}+\frac{1}{9^{s}}+\frac{1}{15^{s}}+\frac{1}{21^{s}}+\frac{1}{27^{s}}+\ldots$$

These denominators are all multiples of $3^{s}$, but not all multiples of $3^{s}$ are here. Some like $6^{s}$ and $12^{s}$ were removed in the previous step. Subtracting these from the previous series leaves terms with denominators that are not multiples of 2 or 3.

$$(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{5^{s}}+\frac{1}{7^{s}}+\frac{1}{11^{s}}+\frac{1}{13^{s}}+\ldots$$

We can't remove terms with multiples of $4^{s}$ because they were sieved out when we removed multiples of $2^{s}$. The next useful step is to remove multiples of $5^{s}$.

Repeating this process leaves terms with denominators that are not multiples of 2, 3 or 5.

$$(1-\frac{1}{5^{s}})\cdot(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{7^{s}}+\frac{1}{11^{s}}+\frac{1}{13^{s}}+\ldots$$

Again, we can't remove multiples of $6^{s}$ because they were already removed. If we repeat this several times, we'll see we can only remove multiples of successive primes. We'll also see that after each removal, the very first term 1 always survives. 

If we kept going, we'd end up with an infinite product on the left, and only 1 on the right.

$$\ldots\cdot(1-\frac{1}{11^{s}})\cdot(1-\frac{1}{7^{s}})(1-\frac{1}{5^{s}})\cdot(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1$$

We can rearrange this to isolate $\zeta(s)$. 

$$\zeta(s)=\prod_{p}(1-\frac{1}{p^{s}})^{-1}$$

The symbol $\prod$ means product, just like $\sum$ means sum.

Euler's Product Formula

We've arrived at Euler's product formula.

$$\boxed{\sum_{n}\frac{1}{n^{s}}=\prod_{p}(1-\frac{1}{p^{s}})^{-1}}$$

The product of $(1-\frac{1}{p^{s}})^{-1}$ over all primes $p$ is the sum of $\frac{1}{n^{s}}$ over all positive integers $n$, as long as we remember to keep $s>1$.

To say this result is amazing would not be an exaggeration. It reveals a deep connection between the primes and the ordinary counting numbers, a connection that doesn't appear too complicated at first sight. 

The Prime Number Theorem

We’ve just seen experimental evidence that $n/\ln(n)$ approximates $\pi(n)$ fairly well. Although the error itself grows as $n \rightarrow \infty$, the proportional error gets ever smaller.

Let’s write that out.

$$ \lim_{n \rightarrow \infty} \frac{\pi(n) - n/\ln(n)}{\pi(n)} = 0 $$

Rearranging this gives us the following.

$$ \lim_{n \rightarrow \infty} \frac{\pi(n)}{n/\ln(n)} = 1 $$

This says the ratio of $\pi(n)$ and the approximation $n/ ln(n)$ tends to 1 as $n \rightarrow \infty$. And this is precisely what the prime number theorem says.

$$ \pi(n) \sim n/ \ln(n) $$

The symbol $\sim$ says that both sides are asymptotically equivalent. For example, $f(n) \sim g(n)$ means $f(n)/g(n) = 1$ as $n \rightarrow \infty$.

Video: [youtube]

Slides: [pdf]

Asymptotic Equivalence

Some examples of asymptotic equivalence will help clarify its meaning.

If $f(x)=x^2 + x$ and $g(x)=x^2$, then $f \sim g$. Both $f$ and $g$ have the same dominant term $x^2$.

$$ \lim_{x\rightarrow\infty} \frac{x^2+x}{x^2} = \lim_{x\rightarrow\infty}1+\frac{1}{x}=1 $$

Swapping $f$ and $g$ doesn't break asymptotic equivalence, $g \sim f$.

$$ \lim_{x\rightarrow\infty} \frac{x^2}{x^2+x}=\lim_{x\rightarrow\infty}\frac{1}{1+\frac{1}{x}}=1 $$

However, if $f(x)=x^{3}$ and $g(x)=x^{2}$, then $f$ and $g$ are not asymptotically equivalent because the ratio $f/g$ tends to $x$, not 1. 

If we know that $f \sim g$ and $g \sim h$, then we can also say $f \sim h$. This property, called transitivity, is familiar from normal equality.

What About li(n)?

Gauss' second approximation $\text{li}(n)$ appeared to be a better approximation for $\pi(n)$. You'll find the prime number theorem is sometimes expressed using the logarithmic integral.

$$ \pi(n) \sim \text{li}(n) $$

Surely the prime number theorem must be about one of the approximations, not both? The only solution is for both approximations to be asymptotically equivalent. Let's see that this is indeed the case.

Let's set $f(n)=\frac{{n}}{\ln(n)}$ and $g(n)=\int_{0}^{n}\frac{1}{\ln(x)}dx$. 

To show $f \sim g$ we need to find the limit of $f(n)/g(n)$ as $n\rightarrow\infty$ and confirm it is 1. Sadly, both $f(n)$ and $g(n)$ become infinitely large as $n\rightarrow\infty$, so the ratio is undefined. 

When this happens, we usually try l'Hopital's rule as an alternative way to find the limit.

$$ \lim_{n\rightarrow\infty}\frac{f(n)}{g(n)}=\lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)} $$

It's fairly easy to work out $f'(n)=\frac{\ln(n)-1}{\ln^{2}(n)}$, and $g'(n)=\frac{1}{\ln(n)}$ pops out of the definition of $\text{li}(n)$. 

$$ \begin{align} \lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)} &= \lim_{n\rightarrow\infty}\frac{(\ln(n)-1)\ln(n)}{\ln^{2}(n)} \\ &= \lim_{n\rightarrow\infty}1-\frac{1}{\ln(n)} \\ &= 1 \end{align} $$

So the prime number theorem can refer to either of the two approximations, $n/ln(n)$ and $\text{li}(n)$, because they are asymptotically equivalent.

What Does The Prime Number Theorem Really Say?

The prime number theorem says that $\pi(n)$ grows in a way that is asymptotically equivalent to functions like $n/\ln(n)$ and $\text{li}(n)$. 

It doesn't say that these are the only or best functions for approximating $\pi(n)$, which leaves open the intriguing possibility of other functions that are even better than $\text{li}(n)$.

Bertrand's Postulate

The prime number theorem, even if it looks imprecise, is rather useful. Let's look at a particularly simple example.

In 1845 Bertrand proposed that there is at least one prime between a counting number and its double, $n<p<2n$.

Let's compare the number of primes up to $2x$, with the number of primes up to $x$.

$$ \frac{\pi(2x)}{\pi(x)} \sim \frac{2x}{\ln(2x)} \cdot \frac{\ln(x)}{x} \sim 2 $$

This shows that for sufficiently large $x$, between $x$ and $2x$, there are $x/\ln(x)$ primes. This is a stronger statement than Bertrand's postulate which merely suggests there is at least one prime.

There are proofs that don't require $x$ to be sufficiently large, but they're nowhere near as simple as this deduction from the prime number theorem.

Historical References For $\pi(n)$

Gauss, 1791

Gauss’ 1791 ‘Some Asymptotic Laws Of Number Theory’ can be found in volume 10 of his collected works. In it he presents his approximation for $\pi(n)$.

$$ \frac{a}{la} $$

Today, this would be written as $n/ \ln(n)$.

Gauss’ 1971 Some Asymptotic Laws Of Number Theory.

Source: http://resolver.sub.uni-goettingen.de/purl?PPN236018647

Legendre, 1797

Legendre in his first edition of ‘Essai Sur La Theorie Des Nombres’ presented his approximation.

$$ \frac{a}{A \text{log}(a)+B} $$

The logarithm is the natural $\ln(a)$. In his 1808 second edition he quantifies the constants.

$$ \frac{x}{\text{log}(x) - 1.08366} $$

Legendre’s 1797 Essai Sur La Theorie Des Nombres.

Source: https://gallica.bnf.fr/ark:/12148/btv1b8626880r/f55. image

Gauss, 1849

Gauss wrote a letter to astronomer Encke dated Decemer 24th 1849, in which he first presents an integral form of a prime counting function. He states this is based on work he started in 1792 or 1793.

Gauss uses the following expression.

$$ \int \frac{dn}{\text{log}n} $$

Today this would be written as the logarithmic integral function.

$$ \int_0^n \frac{1}{\ln(x)} dx $$

First page of Gauss’ 1849 letter to Encke.

Source: https://gauss.adw-goe.de/handle/gauss/199

Experimental Maths - The Distribution Of Primes

Given primes are so resistant to encoding into a simple generating formula, let’s take a detour and try a different approach, experimental mathematics.

Video: [youtube]

Slides: [pdf]

Number of Primes Up To A Number

We showed that primes don’t run out as we explore larger and larger numbers. We also showed they don’t thin out as quickly as the squares. So how quickly do they thin out?

One way to explore this is to keep a count of the number of primes as we progress along the whole numbers.

The expression $\pi(n)$ has become an abbreviation for ‘the number of primes up to, and including, n’. For example, $\pi(5) = 3$ because there are 3 primes up to, and including, 5. The next number 6 is not prime, so $\pi(6)$ remains 3. The use of the symbol $\pi$ can be confusing at first.

The chart below shows $\pi(n)$ for $n$ up to 100. A fairly smooth curve seems to be emerging. This is slightly unexpected because the primes appear to be randomly placed amongst the numbers. The curve suggests the primes are governed by some kind of constraint. It wouldn’t be too adventurous to say the curve looks logarithmic, like $\ln(n)$. (Click to enlarge)

$\pi(n)$ for $n$ from 1 to 100.

Rather Good Approximations for $\pi(n)$

Gauss, one the most prolific mathematicians in history, was the first to find an expression that approximates $\pi(n)$ fairly well. He was aged about 15 at the time.

$$ π(n) \approx \frac{n}{\ln(n)} $$

The expression is surprisingly simple. It is worth pondering on what hidden pattern in the primes is captured by the natural logarithm $\ln(n)$.

Just a year later, Gauss developed a different expression that approximates $\pi(n)$ even more closely.

$$ \pi(n) \approx \int_0^n \frac{1}{\ln(x)} dx $$

At first glance, this logarithmic integral function, shortened to $\text{li}(n)$, appears to be a continuous form of the first approximation. 

The chart below shows a comparison of these approximations with the actual $\pi(n)$ for $n$ all the way up to 10,000. It's clear the logarithmic integral function is much closer to the actual prime counts.

Comparing $\text{li}(n)$ and $n/\ln(n)$ with $\pi(n)$.

Proportional Error

Looking again at the previous chart, the prime counting approximation $n/\ln(n)$ appears to be diverging away from the true prime count $\pi(n)$ as $n$ gets larger. That is, the error appears to be getting ever larger. If we looked at the numbers, we’d also see $\text{li}(n)$ diverging away from $\pi(n)$ too. Does this mean the approximations become useless as $n$ gets larger?

The chart below paints a different picture. It shows the error as a proportion of $\pi(n)$. We can see this proportional error becomes smaller as $n$ grows to 10,000. It's also clear that $\text{li}(n)$ has a distinctly smaller proportional error than $n/\ln(n)$.

Proportional errors for $\text{li}(n)$ and $n/\ln(n)$.

There are 1229 primes amongst the first 10,000 whole numbers. The logarithmic integral gives us $\text{li}(10,000)=1246$. The error is just 17, and as a proportion of 1229, an impressively small 0.0138.

If we extended $n$ to even larger values, we'd find the proportional error would fall further towards zero. Perhaps these approximations are correct in the limit $n\rightarrow\infty$?

Prime Density

Let’s look again at those approximations and see if we can interpret their form. The following compares Gauss’ first approximation with a general expression for calculating the mass of a volume of stuff with a given average density.

$$ \begin{align} \text{mass} &= \text{density} \times \text{volume} \\ \\ \pi(n) &\approx \frac{1}{\ln(n)} \times n \end{align} $$

The comparison suggests that $1/ \ln(n)$ is the average density of primes. If true, this would be a remarkable insight into the primes.

We can apply a similar analogy to Gauss’ second approximation too. This time we compare it with another general expression for calculating mass where the density is not assumed to be constant throughout its volume.

$$ \begin{align} \text{mass} &= \int (\text{density}) dv \\ \\ \pi(n) &\approx \int_0^n \frac{1}{\ln(x)} dx \end{align} $$

Again, $1/\ln(x)$ emerges as a more locally accurate density of primes around a number $x$.

It was this density of primes around a number that the young Gauss first noticed as he studied the number of primes in successive ranges of whole numbers, 1-1000, 1001-2000, 2001-3000, and so on.

Nth Prime?

If the density of primes is $1/ \ln(n)$ then we can say average distance between primes is $\ln(n)$. We can then make the short leap to say the $n^{th}$ prime is approximately $n\ln(n)$.

Before we get too excited about having found a simple function for generating the $n^{th}$ prime, this expression is an approximation, based on an average, itself based on Gauss’ approximation for $\pi(n)$.

It’s still interesting to see how well this expression for the $n^{th}$ prime performs. The chart below shows the error in $n\ln(n)$ as a proportion of the actual $n^{th}$ prime. As $n$ increases to 10,000,000, the proportional error falls to approximately 0.1017.

Proportional error of $n\ln(n)$ against the $n^{th}$ prime.

Looking at the graph, it is tempting to conclude the proportional error in $n\ln(n)$ approaches 0.1 as $n \rightarrow \infty$. We should be cautious because the error could be falling to zero very very slowly.

Anecdotal evidence is not mathematical proof, no matter how compelling it looks.

For example, our experiments show $\text{li}(n)$ is always a bit higher than the true $\pi(n)$. If we extended $n$ from 10,000 to 10,000,000 we’d still find $\text{li}(n)$ was always higher. But in 1914 Littlewood proved that $\text{li}(n)$ can become lower than $\pi(n)$ infinitely many times. More recent proofs show this starts to happen somewhere between $10^{19}$ and a mind-blowingly large $10^{316}$.

Imperfect History

The question of who first developed an approximation for $\pi(n)$ is not perfectly clear. Gauss didn’t always publish his work, leaving us to reconstruct history from notes and letters.

In his 1797 book on number theory, Legendre first published a form $n/(A \ln(n) + B)$, which he updated in his 1808 second edition to $n/(\ln(n) − 1.08366)$.

However, in 1849 Gauss wrote a letter to astronomer, and former student, Encke telling him that he had, in ‘1792 or 1793’, developed the logarithmic integral approximation, which he wrote as $\int\frac{dn}{\text{log}n}$. His collected works also reveal that in 1791 he had written about the simpler approximation, $\frac{a}{la}$ as he wrote it.

A separate blog post presents reproductions of the relevant parts of these historical works.

Primes Aren't That Spread Out

We’ve seen there is no limit to the supply of primes. A good question to ask next is how frequently they occur.

One way to explore this is by looking at the sum of their inverses, or reciprocals.

Video: [youtube]

Slides: [pdf]

Infinite Sum Of Reciprocals

The counting numbers $1, 2, 3, 4, \ldots$ are spaced 1 apart. The sum of their inverses is called the harmonic series.

$$ 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \ldots $$

This series is known to diverge, that is, the sum is infinitely large. A separate blog post has an easy short proof.

The square numbers $1, 4, 9, 16, \ldots$ are spaced further apart than the counting numbers.

$$ 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \ldots $$

The sum of their inverses converges.  A separate blog post walks through Euler’s historic and rather adventurous proof showing it converges to $\frac{\pi^2}{6}$.

We can interpret this to mean the squares $n^2$ are so spread out that the terms in the series become small quickly enough for the sum not to become infinitely large.

It is natural to then ask the same question about the primes. Are they so spread out that the infinite sum of their inverses converges too?

Infinite Sum Of Prime Reciprocals

Let’s start by assuming the infinite series of prime reciprocals does in fact converge to a finite sum $S$.

$$ S = \sum_{n=1} \frac{1}{p_n} $$

Because $S$ is finite, and each term is smaller than the previous one, there must be a value of $k$ such that the infinite series after $\frac{1}{p_k}$ sums to less than 1. We can call this sum $x$. 

$$ x = \sum_{n=k+1} \frac{1}{p_n} <1 $$

Let’s build an infinite geometric series based on this $x$.

$$ G = x + x^2 + x^3 + x^4 + \ldots $$

This new series $G$ converges because the ratio between terms $x$ is less than 1.

Let’s think a little more carefully about the terms in $G$. Any term in $G$ will be of the form $\frac{1}{N}$ where $N$ has prime factors $p_{k+1}$ or larger. This is because $x$ was intentionally constructed with primes $p_{k+1}$ and larger.

Now consider a second series F where, in contrast to G, the terms are constructed from all the primes $p_k$ and smaller.

$$ F = \sum_{j=1} \frac{1}{1 + j \cdot (p_1 \cdot p_2 \cdot p_3 \ldots p_k)}$$

Between each term, only $j$ changes. Now let’s look more closely at the expression $1 + j \cdot (p_1 \cdot p_2 \cdot p_3 \ldots p_k)$. This has no prime factors from the range $p_1$ to $p_k$. Since all whole numbers have prime factors, its prime factors must be from the set $p_{k+1}$ and larger.

That means $F$ is a subseries of $G$. That is, the terms of $F$ appear in the terms of $G$.

Now, if we compare the terms of $F$ to the harmonic series, we can test whether $F$ diverges.

We do this with the limit comparison test, which tests what happens to the ratio of terms from each series as they extend to infinity. If the ratio is finite, the series either both converge, or both diverge.

$$ \lim_{j \rightarrow} \frac{1 + j \cdot (p_1 \cdot p_2 \cdot p_3 \ldots p_k)}{j} = p_1 \cdot p_2 \cdot p_3 \ldots p_k $$

The ratio is finite, and since the harmonic series diverges, so does $F$.

Since $F$ diverges, and is a subseries of $G$, then $G$ must also diverge. But we constructed $G$ to converge. This contradiction proves the initial assumption that the infinite series of prime reciprocals converges was wrong.

That $\sum \frac{1}{p_n}$ diverges is a little surprising because our intuition was that primes thin out rather rapidly.

Legendre’s Conjecture

The fact that $\sum \frac{1}{n^2}$ converges suggests the primes are not as sparse as the squares. This leads us to an interesting proposal attributed to Legendre, but actually first published by Desboves in 1855, that there is at least one prime number between two consecutive squares.

$$ n^2 < p < (n+1)^2 $$

This remains a deep mystery of mathematics. Nobody has been able to prove or disprove it.