Another Proof $\sum \frac{1}{p}$ 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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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, $360=(2\cdot5)\cdot(2\cdot3)^{2}$ has a square-free factor of 10, and a square factor of 36. On the other hand, $30=(2\cdot3\cdot5)$ is entirely square-free. 

 

$\sum' 1/k$  Diverges

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

$$\left(\sum_{k<n}\frac{1}{k}\right)\left(\sum_{j<n}\frac{1}{j^{2}}\right)\geq\sum_{m<n}\frac{1}{m}$$

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 $n\rightarrow\infty$, 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 $\sum1/k$ must diverge, a neat result we'll use very soon.

Proof By Contradiction

Let's assume, perhaps incorrectly, the sum of prime reciprocals $\sum1/p$ 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 $n>3$. 

$$\prod_{p<n}(1+\frac{1}{p})\geq\sum_{k<n}\frac{1}{k}$$

Let's put all this together.

$$\exp(\beta)>\exp\left(\sum_{p<n}\frac{1}{p}\right)=\prod_{p<n}\exp(\frac{1}{p})>\prod_{p<n}(1+\frac{1}{p})\geq\sum_{k<n}\frac{1}{k}$$

This suggests that as $n\rightarrow\infty$ the finite $\exp(\beta)$ is greater than $\sum1/k$, which we showed was divergent. This is clearly a contradiction, so our assumption that $\sum1/p$ converges was wrong. 

$$\boxed{\sum\frac{1}{p}\rightarrow\infty}$$