New answers tagged riemann-zeta-function
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
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
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
$$
\zeta(s)...
- 465
4
votes
Accepted
Bounds of zeta function near $\Re(s)=1$
Heath-Brown (2016) proved that, for any $\varepsilon>0$,
$$\zeta(\sigma+it)\ll_\varepsilon t^{\frac{1}{2}(1-\sigma)^{3/2}+\varepsilon},\qquad 0\leq\sigma\leq 1,\qquad t\geq 1.$$
The exponent $1/2$ ...
- 105k
5
votes
Accepted
Conjectured error term when counting square-free integers
Your guess is correct! It is indeed conjectured that $a=1/4$. A good recent reference is [1]. In particular, it is known that
$$E(x)=\Omega(x^{1/4})$$
and computations have shown
$$|E(x)|<1.12543x^{...
- 465
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
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