All questions with tag [math: analytic-number-theory]

I have recently read an article about the prime number theorem, in which Mathematicians Erdos and Selberg had claimed that proving $\lim \frac{p_n}{p_{n+1}}=1$, where $p_k$ is the $k$th prime, is a very helpful step towards proving the prime number theorem, although I don't know how, primarily because I have not gone through the proof of the the...

We have the following evaluations: $$\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\,\binom {2n}n} = \frac{\pi}{3\sqrt{3}}\\ &\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} = \frac{1}{3}\,\zeta(2)\\ &\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} = -\frac{4}{3}\,\zeta(3)+\frac{\pi\sqrt{3}}{2\cdot 3^2}\,\left(\zeta(2, \tfra...

Is there a simple way to from the prime number theorem? The prime number theorem immediately implies Bertrand's postulate for sufficiently large $n$, but it fails to establish a base case (the linked proof on Wikipedia explicitly gives the base case $n \ge 468$). In the other direction, Bertrand's postulate yields $\pi(n) \ge \log_2(n)$ which ...

Let $\psi(x) := \sum_{n\leq x} \Lambda(n)$ where $\Lambda(n)$ is the Von-Mangoldt function. I want to show that if $$ \lim_{x \rightarrow \infty} \frac{\psi(x)}{x} =1 $$ then also $$\lim_{x\rightarrow \infty} \frac{\pi(x) \log x }{x}=1.$$ I tried to play a little bit with $\psi$, what I want to show is that: $$\left| \frac{\pi(x) \log x}{x} -1 \...

What is the asymptotically slowest growing function $f(x)$, such that there exists constants $a$ and $b$, such that for all $x>a$, there is always a prime between $x$ and $x+bf(x)$? $f(x)=x$ works, does $\sqrt{x}$ work, $\log(x)$, or $\log\log(x)$?

I am feeling a bit slow today. In Analytic Number Theory it is usual to express asymptotic bounds by specifying the relation of the constant to a specific variable, i.e. $\log n \ll_\epsilon n^\epsilon$ which means that $\log n \leqslant C_\epsilon n^\epsilon$ for sufficiently large $n$, where the constant $C_\epsilon$ depends only on the consta...

Let d (n) be the variety of integers much less after that n which has a weird variety of prime variables (2,3,5,7,8,11,12,13,17,18 ). Just how to confirm d (n)/ n have a restriction 1/2? Is there for all m an n such that $|n-2d(n)|>m$?

How can I confirm the adhering to identity: $$\sum_{k=1}^{n}{\sigma_{\ 0} (k^2)} = \sum_{k=1}^{n}{\left\lfloor \frac{n}{k}\right\rfloor \ 2^{\omega(k)}}$$ where $\omega(k)$ is the variety of distinctive prime divisors of $k$.

I'm trying to understand a proof in Chandrasekharan's Introduction to Analytic Number Theory. Specifically, the proof of the lemma on p.118 before Dirichlet's theorem on primes in arithmetic progressions. Define $$ Q(s) = \log P(s) $$ for some particular branch of the logarithm for $\sigma > 1$. If $$ Q(s) = \sum_{n=1}^{\infty} \frac{a_...

The Riemann $\zeta$ function plays a substantial duty in number theory and also is specified by $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} \qquad \text{ for } \sigma > 1 \text{ and } s= \sigma + it$$ The Riemann hypothesis insists that all the non - unimportant absolutely nos of the $\zeta$ function push the line $\text{Re}(s) = \frac...

Background: I was trying to derive an asymptotic formula for the following: $$\sum_{m\leqslant n}\sum_{k\leqslant m}(m\ \mathrm{mod}\ k),$$ which I think I succeeded in doing (I will skip some steps below to come to my question sooner). We have \begin{align} \sum_{m\leqslant n}\sum_{k\leqslant m}(m\ \mathrm{mod}\ k)&=\sum_{m\leqslant n}\...

Wikipedia states that The Riemann zeta function $\zeta(s)$ is defined for all complex numbers $s \neq 1$. It has zeros at the negative even integers (i.e. at $s = −2, −4, −6, ...)$. These are called the trivial zeros. The Riemann hypothesis is concerned with the non-trivial zeros, and states that: The real part of any non-trivial zero of the Ri...

The Divisor summatory function is a function that is an amount over the divisor function. $$D(x)=\sum_{n\le x} d(n) = 2 \sum_{k=1}^u \lfloor\frac{x}{k}\rfloor - u^2, \;\;\text{with}\; u = \lfloor \sqrt{x}\rfloor$$ http://en.wikipedia.org/wiki/Divisor_summatory_function#Dirichlet.27s_divisor_problemI am seeking a formula or a reliable algorithm ...

How to show if $\chi_{0}$ is the unimportant $\text{Dirichlet Character}$ after that $$\lim_{\xi \to 1} (\xi -1) L(\xi,\chi_{0}) = \frac{\Phi(N)}{N}$$ where $\Phi$ is the $\text{Euler's Totient}$.

What is the asymptotic order of numbers divisible by no tops other than those of the kind $an+b$ ($a$, $b$ dealt with)? Undoubtedly (with the exception of the unimportant instances) they are of order purely in between that of he tops and also of all numbers.

Here is a relevant - and longstanding, I'm told - conjecture. Let $f \in \mathbb{Z}[x]$ be irreducible and of degree > 1. Set $E_p = \{x/p \: | \: 0 \leq x < p, f(x) \equiv 0 \: (p) \}$ = normalised least positive residues of zeros of $f$ in $\mathbb{F_p}$ and $E = \bigcup_{p} E_p \subset [0,1]$ Conjecture: $E$ is equidistributed...

For any kind of prime, what percent of the square - free numbers has that prime as a prime variable?

Let $F(x) \in \mathbb{Z}[x]$ and $$ \xi(s) = \sum^\infty_{n=1}g(n)n^{-s} $$ be the Dirichlet series associated an arithmetic function $g(n)$. Define a new Dirichlet series $$ \xi_F(s) = \sum^\infty_{n=1}g(n)F(n)^{-s}. $$ I call $\xi_F(s)$ the Dirichlet series obtained from $\xi(s)$ by shifting by $F(x)$. (Maybe this is known by another na...

Given the j-function $j(\tau)$, $j(\tau) = 1728J(\tau)$, where $J(\tau)$ is Klein’s absolute invariant, the Dedekind eta function $\eta(\tau)$, and the following Eisenstein series, $\begin{align} E_4 (q) &= 1+240\sum_{n=1}^{\infty} \frac{n^3q^n}{1-q^n}\\[1.5mm] E_6 (q) &= 1-504\sum_{n=1}^{\infty} \frac{n^5q^n}{1-q^n}\\[1.5mm] ...

I suggest the Laurent collection at $s=1$. I intend to do it by confirming $\displaystyle \int_0^\infty \frac{2t}{(t^2+1)(e^{\pi t}+1)} dt = \ln 2 - \gamma$, based upon the indispensable formula given up Wikipedia. Yet I can not address this indispensable other than by utilizing Mathematica. Attempted intricate analytic means yet no good luck....