4 Elementary Corollaries
For any arithmetic function \(f\) and real number \(x\), one has
and
Straightforward.
One has
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.
We have
Exponentiate Theorem 17.
Let \(\pi (x)\) denote the number of primes up to \(x\).
One has
as \(x \to \infty \).
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.
One has
as \(x \to \infty \).
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.
Let \(p_n\) denote the \(n^{th}\) prime.
One has
as \(n \to \infty \).
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 \(n\).
We have \(p_{n+1} - p_n = o(p_n)\) as \(n \to \infty \).
Easy consequence of preceding proposition.
For every \(\epsilon {\gt}0\), there is a prime between \(x\) and \((1+\epsilon )x\) for all sufficiently large \(x\).
Use Corollary 9 to show that \(\pi ((1+\epsilon )x) - \pi (x)\) goes to infinity as \(x \to \infty \).
We have \(|\sum _{n \leq x} \frac{\mu (n)}{n}| \leq 1\).
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.
We have \(\sum _{n \leq x} \mu (n) = o(x)\).
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 3 we conclude that
and hence
From Stirling’s formula one has
thus
and thus
Sending \(\epsilon \to 0\) we obtain the claim.
We have \(\sum _{n \leq x} \lambda (n) = o(x)\).
From the identity
and summing, we have
For any \(\epsilon {\gt}0\), we have from Proposition 4 that
and hence on summing in \(d\)
Sending \(\epsilon \to 0\) we obtain the claim.
We have \(\sum _{n \leq x} \mu (n)/n = o(1)\).
As in the proof of Theorem 3, 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 4 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.