13
votes
Accepted
Difficulty with "A new elementary proof of the Prime Number Theorem" by Richter
The proof of Lemma 3.4 uses the constant $K=[\epsilon^{-3}]$ which should be your clue. After the definition of $K$ the author covers $(8^t,8^{t+\epsilon})$ (which contains at least $\epsilon 8^n/(2n)$...
- 2,040
12
votes
Accepted
Can one show that $(-1)^{n-1} {(1/\zeta)}^{(n)}(x) >0$ for all real $x>1$?
In other words, you ask whether the function $f(x):=1-1/\zeta(1+x)$ is completely monotonic on
$[0,+\infty)$. We have $f(x)=\sum_{n>1} -\mu(n)/n^{1+x}=\int e^{-xt}d\lambda(t)$, where $\lambda=\sum_{...
- 109k
12
votes
Accepted
Are all integers up to $x$ but possibly $O_{\varepsilon}(x^{\varepsilon})$ the sum of $a$ squares and $b$ primes with $a+b\leq 3$?
Linnik (1960) proved that every sufficiently large positive integer is the sum of a prime and two squares, confirming a conjecture of Hardy-Littlewood (1923). Moreover, he gave an asymptotic formula ...
- 105k
7
votes
Residue of Dirichlet series at $s = 1$
Yes, the conclusion follows. Indeed, fix any $\varepsilon>0$. For $t>0$ sufficiently large, the sum
$$S(t):=\sum_{n\leq t}a_n$$
satisfies
$$|S(t)-Rt|\leq\varepsilon t.$$ Hence for any $s>1$, ...
- 105k
7
votes
Accepted
Existence of Finite Amicable Groups
Suppose that $G$ and $H$ are nontrivial finite amicable groups and let $\mathscr{G}$ and $\mathscr{H}$ be the nontrivial proper subgroups of $G$ and $H$, respectively. Then $1 \subsetneq \bigoplus_{S ...
- 716
6
votes
Can one show that $(-1)^{n-1} {(1/\zeta)}^{(n)}(x) >0$ for all real $x>1$?
Here is a more pedestrian version of Fedor Petrov's argument: We set $\lambda=1/\log6$ and claim that $(-1)^n(1/\zeta)^{(n)}(\lambda n)>0$ if $n$ is big enough. With $a_k=\log k/k^\lambda$, we have
...
- 22.5k
6
votes
Accepted
Clarification and Proof of Inequality (8.11) in Analytic Number Theory by Iwaniec and Kowalski
The assumptions imply
$$|S_f(N)|\le\sqrt2\frac N{\sqrt q}+2\sqrt q\log q.$$
Indeed, if $q\le2$, this follows from $|S_f(N)|\le N$; for $q\ge3$, we have $\log q>1$, thus
$$\begin{align*}
|S_f(N)|^2&...
- 47.1k
6
votes
Clarification and Proof of Inequality (8.11) in Analytic Number Theory by Iwaniec and Kowalski
The authors seem to claim that if a real quantity $X$ satisfies $X \le N$ and
$$\tag{$*$}\label{483297_star}X^2\le N+\frac{2N^2}{q}+4(N+q)\log q$$
then
$$X\le 2\frac{N}{\sqrt{q}}+\sqrt{q}\log q.$$
...
- 14.6k
6
votes
Accepted
Largest prime factors of integer polynomials
te4 is quite right that this is open. A general comment is that asking whether there exists a subsequence whose lim sup is finite is equivalent to asking whether the lim inf of the original sequence ...
- 148k
6
votes
Accepted
Primes which are safe and Sophie Germain
This is asking for the density of Cunningham chains of the first kind of length three. Take the integer polynomials $f_1(n) = n$, $f_2(n) = 2n+1$ and $f_3(n) = f_2(f_2(n)) = 4n+3$ and apply the (...
- 4,985
5
votes
Accepted
Mellin transform at $0$
By definition, $\tilde{f}(0)$ exists if and only if $\int_0^\infty f(x)/x\,dx$ exists. As $f(x)$ is smooth and supported on $[0,2]$, we have that that
$$|f(x)|=\left|\int_0^x f'(t)\,dt\right|\leq x\...
- 105k
5
votes
Accepted
Conjectured error term when counting square-free integers
Your guess is correct! It is indeed conjectured that $a=1/4$. A good recent reference is [1]. In particular, it is known that
$$E(x)=\Omega(x^{1/4})$$
and computations have shown
$$|E(x)|<1.12543x^{...
- 465
4
votes
Accepted
Bounds of zeta function near $\Re(s)=1$
Heath-Brown (2016) proved that, for any $\varepsilon>0$,
$$\zeta(\sigma+it)\ll_\varepsilon t^{\frac{1}{2}(1-\sigma)^{3/2}+\varepsilon},\qquad 0\leq\sigma\leq 1,\qquad t\geq 1.$$
The exponent $1/2$ ...
- 105k
4
votes
Accepted
Prime number theorem via large sieve type sums
This isn't exactly along the lines you are suggesting, but Hildebrand does have a proof of the Prime Number Theorem which proceeds but estimating $M(x)$ using the large sieve inequality. See:
A. ...
- 12.9k
4
votes
How to prove the identity $L(2,(\frac{\cdot}3))=\frac2{15}\sum\limits_{k=1}^\infty\frac{48^k}{k(2k-1)\binom{4k}{2k}\binom{2k}k}$?
To provide an alternative approach, I'll cross-post my answer from MSE, adapted for the series.
Similarly to what @Nemo obtained in the comments, we can transform the original series into an ...
- 215
4
votes
Accepted
Residue of Dirichlet series at $s = 1$
I think it's a standard exercise in summation by parts if I didn't make a mistake.
We wish to show $$\lim_{\epsilon \to 0^+} \epsilon \sum_{n \ge 1} \frac{a_n}{n^{1+\epsilon}} = R.$$
The sum is the ...
- 1,355
3
votes
First occurrence of formula for $\sum_{n\leq x} \mu(n) \log n$ in terms of $\psi(y)-\lfloor y\rfloor$?
I'm willing to give Landau half-marks for section 155 (chapter 41, part 11, vol II) in his Handbuch: he derives $M(x)=o(x)$ from PNT by noting, in effect, that
$$\sum_{n\leq x} \mu(n) \log n = - \sum_{...
- 20.2k
3
votes
Bounds of zeta function near $\Re(s)=1$
In addition to GH from MO's answer, if one wishes to keep the $(\log|\Im(s)|)^{2/3}$ factor, then there is a very recent improvement due to Bellotti [1]. In particular, Bellotti proved that
$$
\zeta(s)...
- 465
2
votes
Clarification and Proof of Inequality (8.11) in Analytic Number Theory by Iwaniec and Kowalski
After further exploration, I found a related question and discussion on
https://mathoverflow.net/questions/208886/iwaniec-kowalski-exponential-sum-for-quadratic-function?rq=1`
that addressed my ...
- 111
2
votes
Accepted
Bounding a number-theoretic integral
The integral looks something like $$\sum _{n=1}^\infty \frac {\Lambda (n)}{n^c}\int _1^Tt^{1/2-c}\cdot e(t-t\log (X/nt))\cdot dt\hspace {10mm}e(z)=e^{2\pi iz}.$$
The derivative of the phase is ...
- 1,381
2
votes
Possible refinements of the large sieve inequality
Though a bit late --
In general, look at Theorem 2.1 of
https://ramare-olivier.github.io/Maths/Eigenvalues-JTNB.pdf
This improves a bit on the c in (N+cQ^2).
For a generic sifted set, look at Theorem ...
- 341
1
vote
Asymptotic behavior of weighted sums involving the fractional part function
Not an answer but a conjectural answer for the value of $C(m)$ supported by
extensive numerical evidence:
$$C(m)=\dfrac{1}{m-1}\sum_{2\le k\le m-1}\binom{m-1}{k}\zeta(k)+\sum_{2\le k\le m-2}\dfrac{1}{...
- 13.1k
Only top scored, non community-wiki answers of a minimum length are eligible