Straightforward.

From the prime number theorem we already have

so it suffices to show that

Only the terms with \(j \leq \log x / \log 2\) contribute, and each \(j\) contributes at most \(\sqrt{x} \log x\) to the sum, so the left-hand side is \(O( \sqrt{x} \log ^2 x ) = o(x)\) as required.

Exponentiate Theorem 17.

We have the identity

as can be proven by interchanging the sum and integral and using the fundamental theorem of calculus. For any \(\epsilon \), we know from Theorem 17 that there is \(x_\epsilon \) such that \(\sum _{p \leq t} \log p = t + O(\epsilon t)\) for \(t \geq x_\epsilon \), hence for \(x \geq x_\epsilon \)

where the \(O_\epsilon (1)\) term can depend on \(x_\epsilon \) but is independent of \(x\). One can evaluate this after an integration by parts as

for \(x\) large enough, giving the claim.

An integration by parts gives

We have the crude bounds

and

and combining all this we obtain

and the claim then follows from Theorem 18.

Use Corollary 9 to show that for any \(\epsilon {\gt}0\), and for \(x\) sufficiently large, the number of primes up to \((1-\epsilon ) n \log n\) is less than \(n\), and the number of primes up to \((1+\epsilon ) n \log n\) is greater than \(n\).

Easy consequence of preceding proposition.

Use Corollary 9 to show that \(\pi ((1+\epsilon )x) - \pi (x)\) goes to infinity as \(x \to \infty \).

From Möbius inversion \(1_{n=1} = \sum _{d|n} \mu (d)\) and summing we have

for any \(x \geq 1\). Since \(\lfloor \frac{x}{d} \rfloor = \frac{x}{d} - \epsilon _d\) with \(0 \leq \epsilon _d {\lt} 1\) and \(\epsilon _x = 0\), we conclude that

and the claim follows.

From the Dirichlet convolution identity

and summing we obtain

For any \(\epsilon {\gt}0\), we have from the prime number theorem that

(divide into cases depending on whether \(x/d\) is large or small compared to \(\epsilon \)). We conclude that

Applying 4 we conclude that

and hence

From Stirling’s formula one has

thus

and thus

Sending \(\epsilon \to 0\) we obtain the claim.

From the identity

and summing, we have

For any \(\epsilon {\gt}0\), we have from Proposition 5 that

and hence on summing in \(d\)

Sending \(\epsilon \to 0\) we obtain the claim.

As in the proof of Theorem 4, we have

so it will suffice to show that

Let \(N\) be a natural number. It suffices to show that

if \(x\) is large enough depending on \(N\). We can split the left-hand side as the sum of

and

The first term is clearly \(O(x/N)\). For the second term, we can use Theorem 5 and summation by parts (using the fact that \(x/d-j\) is monotone and bounded) to find that

for any given \(j\), so in particular

for all \(j=1,\dots ,N-1\) if \(x\) is large enough depending on \(N\). Summing all the bounds, we obtain the claim.

This is a routine modification of the proof of Theorem 17.