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Assume that $f(n)=o(n^{-2})$. Then $g(n):=n^{-1}\sqrt{f(n)}=o(n^{-2})$, so $f(n)=\Omega(g(n))$, so $\sqrt{f(n)}=\Omega(n^{-1})$, so $f(n)=\Omega(n^{-2})$, a contradiction. This proves that $f(n)=\Omeg …

We have $f(\gamma,\beta)=0$ for every $\gamma>0$ and $\beta\in\mathbb{R}$. Indeed, $1-\gamma^{1/x}$ is asymptotically $(\log\gamma)/x$, and $(\log x)^\beta/x$ tends to zero for any $\beta\in\mathbb{R} …

Yes, these are standard things in analytic number theory. We have $$ \sum_{n=1}^\infty\frac{\tau(n)^k}{n^s}=\zeta(s)^{2^k}F_k(s),\qquad\Re(s)>1,$$ where $F_k(s)$ is an explicit Dirichlet series that c …

This is my third response. I claim that for $0 < t < 1$ we have the uniform bounds $$ \liminf_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\mu}+O(1),$$ $$ \limsup_{n\to\infty}\ nt f_n(t) = \frac{-1}{\log\m …

$F(H)$ and $G(H)$ do not determine the asymptotic growth of the third display. Indeed, consider the following two functions from $\mathbb{Z}_{\geq 0}$ to $\mathbb{Z}_{\geq 0}$: $$ f(x):=\begin{cases}0 …

The statement is true in the stronger form that $$\Pr[\gcd(t, N)>N^{3/4}] < N^{-1/2}.$$ Indeed, the probability that $\gcd(t,N)$ equals a given $k\mid N$ is at most $1/k$. For $k>N^{3/4}$, this is les …

This is an improvement of my previous post. I claim that $$4\pi n-6\pi<f(n)<4\pi n-2\pi.$$ Starting from $$f(n)-2\pi(n-1)=\int_0^\infty \log\left(\frac{1+t}{2}+\frac{1+t}{2}\left(\frac{1-t}{1+t}\right …

You can derive a very precise asymptotic expansion of your quantity by the Selberg-Delange method. I recommend that you adapt, to your situation, the arguments of Section II.6.1 of Tenenbaum: Introd …

Let us introduce the notation $$M(T):=\int_0^T|\zeta(\sigma+it)|^2\,dt.$$ Then $$\int_0^T \frac{|\zeta(\sigma+it)|^2}{\sqrt{1+t^2}} \,dt=\int_0^T\frac{dM(t)}{\sqrt{1+t^2}}=\frac{M(T)}{\sqrt{1+T^2}}+\i …

The sum in question equals \begin{align*}\sum_{p\leq x}\frac{1}{p^{1+2/\log x}}\left(\frac{\log\left(x/p\right)}{\log(x)}\right)^2 &=\sum_{p\leq x}\frac{1}{p}e^{-2\frac{\log p}{\log x}}\left(1-\frac{\ …

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\su …

Your sum can be rewritten as $$ D(N)=\sum_{n=1}^{N-1}d(n)d(N-n),$$ where $N=2c$, and $d(m)$ is the number of divisors of $m$. This is a so-called "binary additive divisor sum", and it has been studied …

See Erick Wong's response here. In particular, Kevin Ford proved (in more precise form) that $$ f(n) = \frac{n}{\log n} \exp\left(O(\log \log \log n)^2\right),$$ whence $f(n)/n$ tends to zero. The sam …

You can find an in-depth answer to your question in this paper of de Reyna and Jeremy. See in particular (65)-(66) along with (30) and Theorem 4.9. See also Theorem 6.2.

The asymptotic formula is true for even dimensions $k\geq 2$. We can prove this by induction on $k$, inspired by Rodrigo's observation on Eisenstein series. The case $k=2$ is classical and addressed …