1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13 ...

We're going to look at how the infinite sum of prime reciprocals behaves. To start off, let's take a closer look at the harmonic series, or the sum of the reciprocals of all numbers: \[\displaystyle\sum_{n=1}^\infty \frac{1}{n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots = \infty \]

We can show that this sum diverges by comparing this to the following Taylor series

ln(11x)=n=1xnn ln\left(\frac{1}{1-x}\right) = \displaystyle\sum_{n=1}^\infty \frac{x^{n}}{n}

Evaluating the limit at x=1x=1

ln()=n=11n= ln(\infty) = \displaystyle\sum_{n=1}^\infty \frac{1}{n} = \infty

If this feels hand-wavy, that's because it is. However this result still turns out to be valid when more careful work is done. But how can this sum tell us anything about the sum of prime reciprocals? To start off, the fundamental theorem of arithmetic tells us that any integer nn can be written uniquely as a product of prime numbers, or n=p1k1p2k2p3k3...n = p_{1}^{k_{1}}p_{2}^{k_{2}}p_{3}^{k_{3}}..., where knk_{n} is either 00 or a positive integer. This uniqueness leads to the next equation

n=11n=p=p1(1+1p+1p2+1p3+)\displaystyle\sum_{n=1}^\infty \frac{1}{n} = \displaystyle\prod_{p=p_{1}}^\infty (1 + \frac{1}{p} + \frac{1}{p^{2}} + \frac{1}{p^{3}} + \cdots)

Essentially, this is a product over all powers of all prime numbers. Thanks to the fundamental theorem of arithmetic, we know that all possible combinations of primes and their powers will lead to all possible whole numbers. Thus, this formula is valid. Using a Taylor series allows us to simplify this expression to

n=0xn=11xx<1 \displaystyle\sum_{n=0}^\infty x^{n} = \frac{1}{1-x} \quad x<1 p=p1(1+1p+(1p)2+(1p)3+)=p=p111p1 \displaystyle\prod_{p=p_{1}}^\infty (1 + \frac{1}{p} + \left(\frac{1}{p}\right)^{2} + \left(\frac{1}{p}\right)^{3} + \cdots) = \displaystyle\prod_{p=p_{1}}^\infty \frac{1}{1-p^{-1}}

Taking the logarithm of the last expression yields

ln(n=11n)=ln(ln())=p=p1ln(11p1) ln\left(\displaystyle\sum_{n=1}^\infty \frac{1}{n}\right) = ln(ln(\infty)) = \displaystyle\sum_{p=p_{1}}^\infty ln\left(\frac{1}{1-p^{-1}}\right)

Using the Taylor series we used in the beginning of this note, we can turn the last equality into

ln(ln())=p=p1(1p+12p2+13p3+)p=p11p ln(ln(\infty)) = \displaystyle\sum_{p=p_{1}}^\infty \left(\frac{1}{p} + \frac{1}{2p^{2}} + \frac{1}{3p^{3}} + \cdots \right) \sim \displaystyle\sum_{p=p_{1}}^\infty \frac{1}{p}

We can ignore the higher-order terms because they are bounded by ζ(n)\zeta(n), which converges to a finite value for n>1n>1. Finally, we get

p=p11p=ln(ln())= \displaystyle\sum_{p=p_{1}}^\infty \frac{1}{p} =ln(ln(\infty)) = \infty

Note by Levi Walker
2 years, 8 months ago

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