## Prime Divergence

Did you know that:

$\sum_{n=prime}^{ \infty}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{prime_n}+...$

is equal infinity? Now, some of you might be thinking, “Oh, I already knew that.” Well, then this post is not for you, move along. Others, though, might think, “hold on, isn’t adding anything up to infinity equals infinity?” Nope because the sum (1/2)+(1/4)+(1/8)+ etc for example is equal to 1. And in fact, all geometric sum (in case you don’t know, it goes like n+n^2+n^3+… etc) with a number less than one converge into a number that is not infinite.

Some others who has learned stuff from second semester calculus class might think “Really? It looks real slow.” Well, it is really slow, but as you shall see, it nevertheless does diverge, and demonstrating this will be the topic of this post.

This post is based on a section in the book “An Imaginary Tale” by Paul Nahin. The book talks about this, and I thought the mathematics was super neat, and so I decided to put it here.

The post is in three parts:

a)I will show a proof originally done by Nicole Oresme (1320-1382) on the divergence of the harmonic series, which is 1+(1/2)+(1/3)+…

b)Then I will change the Riemann zeta function, which is f(x)=1+(1/2^x)+(1/3^x)… into a product of primes, which was done by Leonard Euler (1707-1783).

c)And finally I will use the above results to prove that the sum of all 1/prime does diverge, which was proved by Euler.

The Divergence of the Harmonic Series

This one is really easy:

$\sum_{n=1}^{\infty} \frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$

Now, group those numbers like this:

$S_1=1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+...$

Then form a series that you know is infinite that is similar to the above, but slower:

$S_2=1+\frac{1}{2}+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})+...=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+$

$...=\infty$

It makes sense, right? Since one half plus one half plus one half plus on and on is gonna equal to infinity. Finally:

$S_1>S_2=\infty$

And so the harmonic series is divergent.

Riemann Zeta Function as a Product of Primes

This one is something I thought was really amazing:

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

Then do the sieve of Eratosthenes like this:

$\frac{1}{2^x} \zeta(x)=\frac{1}{2^{x}}+\frac{1}{4^{x}}+\frac{1}{6^{x}}+...$

$\zeta(x)-\frac{1}{2^x} \zeta(x)=(1-\frac{1}{2^x})\zeta=1+\frac{1}{3^{x}}+\frac{1}{5^{x}}+\frac{1}{7^{x}}+...$

See what I did above? I first multiplied the zeta function by the smallest prime number, turning all its numbers into evens, and subtracted the original function with all the evens. Now all that is left is the odds. Now, we do the same, but instead of 2, we do it with the prime afterwards, which is 3, and get all the multiples of threes that are leftover eliminated:

$(1-\frac{1}{2^{x}}) \frac{1}{3^{x}} \zeta(x)=1+\frac{1}{3^{x}}+\frac{1}{9^{x}}+\frac{1}{15^{x}}+...$

$(1-\frac{1}{2^x}) \zeta (x) - (1-\frac{1}{2^x}) \frac{1}{3^x} \zeta(x)=(1-\frac{1}{2^x})(1-\frac{1}{3^x}) \zeta (x)=$

$1+\frac{1}{5^{x}}+\frac{1}{7^{x}}+\frac{1}{11^{x}}+...$

You keep doing that with the prime numbers, and what you will eventually get is:

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

Dividing both sides by the products of prime:

$\zeta(x)=\prod_{p=prime} (1-\frac{1}{p^{x}})^{-1}=\sum_{n=1}^{\infty} \frac{1}{n^{x}}$

And now the final step: Your brain melts after your mind is blown.

The Divergence of the Sum of the Reciprocal of All Primes

Okay, this is the final step of the proof. First of all, notice that zeta(1) is the harmonic series, and so:

$\sum_{n=1}^{\infty} \frac{1}{n}=S_1=\prod_{p=prime} (1-\frac{1}{p})^{-1}$

Taking the logarithm of both sides, and using the fact that log(a*b)=log(a)+log(b) and log(x^a)=a*log(x):

$ln(S_1)=-\sum_{p=prime} ln(1-\frac{1}{p})$

We can turn that natural logarithm in the right hand side into an infinite series using a Taylor series expansion:

$\ln{(1-\frac{1}{p})}=-\frac{1}{p}-\frac{1}{2} \cdot \frac{1}{p^2}-\frac{1}{3} \cdot \frac{1}{p^3}-...$

Pluggin the above back in the infinite sum:

$\ln{S_1}=\sum_{p=prime}(\frac{1}{p}+\frac{1}{2}\frac{1}{p^{2}}+\frac{1}{3}\frac{1}{p^{3}}+...)=\sum_{p=prime}\frac{1}{p}+number$

The reason that all the other terms are some numbers is because for the sum of each term, you can create convergent geometric series that are larger than them. Basically, the logic goes: geometric series=some number>sum of terms of prime numbers. The first one is the one you can’t create a geometric series for because if you try to do it, you form the sum 1+1^2+1^3+… and it diverges, but it doesn’t tell us anything about what happens to the sum of the first term.

What’s left to do is find out what is ln(S1). S1 is the harmonic series, so:

$\ln{\infty}=\infty=\sum_{p=prime}\frac{1}{p}+number$

Because of that, the sum of all 1/prime must be infinite. And I died writing all this. The end.