Sum of Reciprocals of Primes is Divergent
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Theorem
- $\displaystyle \sum_{\substack {p \in \Bbb P \\ p \le n} } \frac 1 p > \ln \left({\ln \left({n}\right)}\right) - \ln \left({\frac {\pi^2} 2}\right)$
- $\displaystyle \lim_{n \to \infty} \left({\ln \left({\ln \left({n}\right)}\right) - \ln \left({\frac {\pi^2} 2}\right)}\right) = + \infty$
Proof
"Zelmerszoetrop has this written up from a final exam last semester, I'll post it up today."
Prime.mover is still looking at his watch.
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Proof of Limit
Observe the following simplification:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \ln \left({\ln \left({n}\right)}\right) - \ln \left({\frac{\pi^2}2}\right)\) | \(=\) | \(\displaystyle \ln \left({\frac {2 \ln n} {\pi^2} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Sum of Logarithms |
Fix $c \in \R$. It suffices to show there exists $N \in \N$, such that:
- $(1):\quad \displaystyle n \ge N \implies \ln \left({\frac {2 \ln n} {\pi^2} }\right) > c$
Proceed as follows:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \ln \left({\frac {2 \ln n} {\pi^2} }\right)\) | \(>\) | \(\displaystyle c\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \frac {2 \ln n} {\pi^2}\) | \(>\) | \(\displaystyle \exp c\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Exponential | ||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \ln n\) | \(>\) | \(\displaystyle \frac {\pi^2 \exp c} 2\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle n\) | \(>\) | \(\displaystyle \exp \left({\frac {\pi^2 \exp c} 2}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Exponential |
Now, obviously, any $N$ with $N > \exp \left({\dfrac {\pi^2 \exp c} 2}\right)$ satisfies condition $(1)$ by Logarithm is Strictly Increasing and Concave.
$\blacksquare$
