Second Chebyshev Function is Increasing
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Theorem
The second Chebyshev function $\psi$ is increasing.
Proof
Let $x \ge y$.
Then:
| \(\ds \map \psi y\) | \(=\) | \(\ds \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le y} \ln p\) | Definition of Second Chebyshev Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop \ge 1} \paren {\sum_{p^k \mathop \le x} \ln p + \sum_{x \mathop < p^k \mathop \le y} \ln p}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p + \sum_{k \mathop \ge 1} \sum_{x \mathop < p^k \mathop \le y} \ln p\) |
From Logarithm is Strictly Increasing:
- $\ln p \ge \ln 2 > 0$
So, we have:
- $\ds \sum_{k \mathop \ge 1} \sum_{x \mathop < p^k \mathop \le y} \ln p \ge 0$
so:
| \(\ds \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p + \sum_{k \mathop \ge 1} \sum_{x \mathop < p^k \mathop \le y} \ln p\) | \(\ge\) | \(\ds \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \psi x\) | Definition of Second Chebyshev Function |
So if $x \le y$, then:
- $\map \psi x \le \map \psi y$
so:
- $\psi$ is increasing.
$\blacksquare$