## Mertens’ first theorem:

Mertens’ first theorem may be expressed as the equation:

$$\sum_{n=1}^{\pi(N)} \frac{\ln p_n}{p_n} = \ln N$$

Mertens’ first theorem may be understood as a precursor of the Prime Number Theorem and it may be easily deduced from the PNT but it also raises important questions about the prime numbers from an information-theoretic perspective which I shall address in a future article.

## Proof:

If we start with the Von Mangoldt function $$\Lambda(\cdot)$$:

$$\log n = \sum_{p^k | n} \Lambda(p^k) = \sum_{d|n} \Lambda(d)$$

we may deduce that:

$$\sum_{n \leq x} \log n = \sum_{d \leq x} \Lambda(d) \cdot \lfloor \frac{x}{d} \rfloor \approx x \sum_{d \leq x} \frac{\Lambda(d)}{d}$$

Meanwhile, due to Stirling we have:

$$\sum_{n \leq x} \log n \approx x \cdot \log x + x$$

By combining (3) and (4) we have:

$$\sum_{n \leq x} \log n \approx \sum_{d \leq x} \frac{\Lambda(d)}{d}$$

and we’ll note that:

$$\sum_{d \leq x} \frac{\Lambda(d)}{d} = \sum_{k=1}^{\lfloor \log_2 (x) \rfloor} \sum_{p \leq x^{1/k}} \frac{\log p}{p^k}$$

which simplifies to:

$$\sum_{d \leq x} \frac{\Lambda(d)}{d} = \sum_{p \leq x} \frac{\log p}{p} + \mathcal{O}(\sum_{p \leq x^{1/2}} \frac{\log p}{p^2}\big(\frac{1}{1-p})\big) = \sum_{p \leq x} \frac{\log p}{p} + \mathcal{O}(1)$$

The last argument is correct as we have:

$$\sum_{p \leq x^{1/2}} \frac{\log p}{p^2}\big(\frac{1}{1-p}\big) < \sum_{n=2}^{\infty} \frac{\log n}{n^2} = \frac{\pi^2}{6}\cdot (12 \cdot \ln(A) - \gamma - \ln(2 \cdot \pi))$$

## References:

1. Mark B. Villarino. Mertens’ Proof of Mertens’ Theorem. 2005.

2. Sondow, Jonathan and Weisstein, Eric W. “Mertens Theorem.” From MathWorld–A Wolfram Web Resource. https://mathworld.wolfram.com/MertensTheorem.html