Another Proof ∑1p Diverges

Ivan Niven authored a very short and rather fun proof that the sum of inverse primes diverges. 

The proof was published in 1971 in the American Mathematical Monthly vol 78, issue 3, 272-273. It is available as open access here: https://www.tandfonline.com

Although the proof is short, it moves at a pace that might leave some behind. Here we'll work through the proof and explain each step more fully.

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Slides: [pdf]

Square-Free Numbers

The proof starts with the idea of square-free positive integers. 

We can write any counting number m as a unique product of a square $j^2$ and square-free factor $k$. 

$$m=k\cdot j^2$$

Remember that any integer is a unique product of primes. We can split these primes into two groups, one group with primes raised to an even power, which together can be written as a square, and the other group with primes not raised to any power. 

For example, $360=(2\cdot 5)\cdot (2\cdot 3{)}^2$ has a square-free factor of 10, and a square factor of 36. On the other hand, $30=(2\cdot 3\cdot 5)$ is entirely square-free. 

 

${\sum }^{\prime }1/k$  Diverges

Let's look at the following inequality, where $k$ are the square-free integers less than $n$.

$$\left(\sum _{k<n}\frac{1}{k}\right)\left(\sum _{j<n}\frac{1}{j^2}\right)\ge \sum _{m<n}\frac{1}{m}$$

The inequality is true because multiplying out the two series would give us not just the terms $1/m$, but also many more. The two sides are only equal when $n=2$. 

As $n\to \mathrm{\infty }$, the right hand side becomes the harmonic series which we know diverges. We also know the second sum converges to ${\pi }^2/6$. That means the sum over square-free integers $\sum 1/k$ must diverge, a neat result we'll use very soon.

Proof By Contradiction

Let's assume, perhaps incorrectly, the sum of prime reciprocals $\sum 1/p$ converges to a finite \beta. 

The partial sum is less than the full sum, $\sum _{p<n}1/p<\beta$, so we can write the following. 

$$\exp (\beta )>\exp \left(\sum _{p<n}\frac{1}{p}\right)=\prod _{p<n}\exp (\frac{1}{p})$$

We can also truncate the Taylor series for $e^x$ to say $e^x>1+x$. Applying this to $\exp (1/p)$ lets us write the following.

$$\prod _{p<n}\exp (\frac{1}{p})>\prod _{p<n}(1+\frac{1}{p})$$

Multiplying out that product would give a series with terms $1/k$ where each $k$ is square-free. This is because each prime contributes to any $k$ at most once. We'd also end up with more terms than are in $\sum _{p<n}1/k$ for $n>3$. 

$$\prod _{p<n}(1+\frac{1}{p})\ge \sum _{k<n}\frac{1}{k}$$

Let's put all this together.

$$\exp (\beta )>\exp \left(\sum _{p<n}\frac{1}{p}\right)=\prod _{p<n}\exp (\frac{1}{p})>\prod _{p<n}(1+\frac{1}{p})\ge \sum _{k<n}\frac{1}{k}$$

This suggests that as $n\to \mathrm{\infty }$ the finite $\exp (\beta )$ is greater than $\sum 1/k$, which we showed was divergent. This is clearly a contradiction, so our assumption that $\sum 1/p$ converges was wrong. 

$$\boxed{{\displaystyle \sum \frac{1}{p}\to \mathrm{\infty }}}$$

Integral Comparison Tests

Understanding the behaviour of continuous functions is often easier than discrete functions. We can gain insights into discrete sums like $\sum \frac{1}{x}$ by exploring the related continuous integral $\int \frac{1}{x}dx$.

This is a simple but powerful technique used a lot in number theory, and worth becoming familiar with.

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Slides: [pdf]

Lower & Upper Bounds For The Growth Of $\sum 1/n$

The picture below shows a graph of $y=\frac{1}{x}$, together with rectangles representing the fractions $\frac{1}{n}$.

Lower Bound

If we consider the range $1\le x\le 4$ we can see the area of the three taller rectangles $1+\frac{1}{2}+\frac{1}{3}$ is greater than the area under the curve ${\int }_1^4\frac{1}{x}dx$. By extending the range to $n$ we can make a general observation.

$$\sum _1^n\frac{1}{x}>{\int }_1^{n+1}\frac{1}{x}dx$$

The integral has an upper limit of $n+1$ because the width of the last rectangle extends from $x=n$ to $x=n+1$. We can perform the integral to simplify the expression.

$$\boxed{{\displaystyle \sum _1^n\frac{1}{x}>\ln (n+1)}}$$

This is a rather nice lower bound on the growth of the harmonic series.

Upper Bound

Let's now look at the shorter rectangles. In the range $1\le x\le 4$ we can see the area of the three shorter rectangles $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$ is less than the area under the curve ${\int }_1^4\frac{1}{x}dx$. Again, by extending the range to n we can make a general observation.

$$\sum _2^n\frac{1}{x}<{\int }_1^n\frac{1}{x}dx$$

The harmonic sum starts at 2 because this time we're looking at rectangles extending to the left of a given $x$. We can adjust the limit of the sum using $\sum _1^n\frac{1}{x}=1+\sum _2^n\frac{1}{x}$.

$$\sum _1^n\frac{1}{x}-1<{\int }_1^n\frac{1}{x}dx$$

Again, we can perform the integral.

$$\boxed{{\displaystyle \sum _1^n\frac{1}{x}<\ln (n)+1}}$$

This is a nice upper bound to the growth of the harmonic series.

Convergence Of $\zeta (s)=\sum 1/n^s$

The picture below shows a graph of $y=\frac{1}{x^s}$, together with rectangles representing the fractions $\frac{1}{x^s}$. The shape of the graph assumes $s>0$. If $s$ was $\le 0$ then it is easy to see $\sum 1/n^s$ would diverge because each term would be $\ge 1$.

If we consider the range $1\le x\le 4$ we can see the area of the three shorter pink rectangles is less than the area under the curve ${\int }_1^4\frac{1}{x^s}dx$. By extending the range to $k$ we can make a general observation.

$$\sum _2^k\frac{1}{x^s}<{\int }_1^k\frac{1}{x^s}dx$$

The sum starts at 2 because we're looking at rectangles extending to the left of a given x. We can adjust the limit of the sum using $\sum _1^k\frac{1}{x^s}=1+\sum _2^k\frac{1}{x^s}$.

$$\sum _1^k\frac{1}{x^s}-1<{\int }_1^k\frac{1}{x^s}dx$$

The integral is easily evaluated.

$$\sum _1^k\frac{1}{x^s}<\frac{k^{1-s}-1}{1-s}+1$$

As $k\to \mathrm{\infty }$, the right hand side only converges when $s>1$. Because it is less than the right hand side, the sum $\sum 1/x^s$ also converges when $s>1$. We haven't yet ruled out the possibility the sum might also converge for some $s\le 1$.

If we now consider the three taller rectangles $1+\frac{1}{2^s}+\frac{1}{3^s}$ in the range $1\le x\le 4$, we can see their area is greater than the area under the curve ${\int }_1^4\frac{1}{x^s}dx$. By extending the range to $k$, we can make a general observation.

$$\sum _1^k\frac{1}{x^s}>{\int }_1^{k+1}\frac{1}{x^s}dx$$

The integral has an upper limit of $k+1$ because the width of the last rectangle extends from $x=k$ to $x=k+1$. We can perform the integral to simplify the expression.

$$\sum _1^k\frac{1}{x^s}>\frac{(k+1{)}^{1-s}-1}{1-s}$$

As $k\to \mathrm{\infty }$, the right hand side diverges when $s\le 1$. Because it is greater than the right hand side, the sum $\sum 1/x^s$ also diverges when $s\le 1$. We have now ruled out the possibility the sum might converge for some $s\le 1$.

$$\boxed{{\displaystyle \zeta (s)=\sum 1/n^s\text{ only converges for }s>1}}$$

We can go further. The two inequalities together provide a lower and upper bound for the zeta function.

$$\boxed{{\displaystyle \frac{1}{s-1}<\zeta (s)<\frac{1}{s-1}+1}}$$