\sum_{n \leq x} lo g n = \sum_{d \leq x} Λ (d) ⌊ \frac{x}{d} ⌋ = x \sum_{d \leq x} \frac{Λ ( d )}{d} + O (ψ (x)) .

By the Stirling formula, the left side is

lo g (⌊ x ⌋!) = x lo g x - x + O (lo g x)

, while the assumed Chebyshev estimate gives

ψ (x) = O (x)

. Hence

\sum_{n \leq x} \frac{Λ ( n )}{n} = lo g x + O (1) .

The contribution of proper prime powers is bounded uniformly:

\sum_{p^{k} \leq x k \geq 2} \frac{l o g p}{p ^{k}} \leq \sum_{p} \frac{l o g p}{p ( p - 1 )} < \infty.

Removing it leaves

p \leq x \sum \frac{lo g p}{p} = lo g x + O (1) .

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