# Understanding Wright's proof of Landau's theorem

I'm reading Wright's A simple proof of a theorem of Landau in which the core argument is a proof by induction and I find myself stuck on a major point. I must be misunderstanding notation or something else, because what I understand is clearly wrong.

For the base case, he writes:

But, for $k=1,$ (6) is equivalent to (2).

where (2) is a form of the Prime Number Theorem, $$\vartheta(x)=\sum_{p\le x}\log p\sim x$$

and (6) is $$\phi_k(x)=o\left\{(\log\log x)^{k-1}\right\}$$

The relevant definitions: \begin{align} \phi_k(x)&=\vartheta_k(x)-kx\Omega_{k-1}(x)\\ \vartheta_k(x)&=\sum_{p_1\cdots p_k\le x}\log(p_1\cdots p_k)\\ \Omega_0(x)&=1\\ \Omega_k(x)&=\sum_{p_1\cdots p_k\le x}\frac{1}{p_1\cdots p_k}\text{ for }k>0\\ \end{align} (where both $\vartheta_k$ and $\Omega_k$ count their sums with multiplicity) and so $$\phi_1(x)=\vartheta(x)-x$$ but $|\psi(x)-x|\ne o(\sqrt x\log\log\log x)$ and $\psi(x)-\vartheta(x)=O(\sqrt x)$ hence $|\vartheta(x)-x|\ne o(\sqrt x\log\log\log x),$ and so $\phi_1(x)=\vartheta(x)-x\ne o(1)$ as required. (In any case it can't be $o(1)$ since there are arbitrarily long gaps in the primes.) What am I missing?

• Reading the title I thought it was about Landau's Theorem on Dirichlet series. Perhaps you should give a more thorough background for what you exactly want. – Timbuc Jul 16 '15 at 18:22
• @Timbuc: the proof seems to assume that theta(x) - x goes to 0, but it doesn't. What don't I get? – Charles Jul 16 '15 at 18:42
• You didn't define $\Omega_k(x)$. – Eric Naslund Jul 16 '15 at 20:28
• @EricNaslund: Yes, only $\Omega_0(x),$ because that's all that's used in the question. I can copy the full definition for you if you're interested. – Charles Jul 16 '15 at 20:33
• @Charles: Well, yes, it's definitely relevant since this looks like a typo. – Eric Naslund Jul 16 '15 at 20:42

It looks like a typo to me, and that equation $(6)$ should read $$\phi_k(x)=o\left(x\left(\log \log x\right)^{k-1}\right).$$
Knowing how the author defines $\Omega_{k-1}(x)$ for other values of $k$ would help confirm this.