All Questions
7 questions
1
vote
0
answers
100
views
Prove or disprove that $|(1/\zeta)^{(n)}(x)| \leq \frac{n!}{(x-\frac{1}{2})}$ for all real $x>1$
$|(1/\zeta)^{(n)}(x)| \leq \frac{n!}{(x-\frac{1}{2})}$ for all real $x>1$.
I had this conjecture for a long time. I tried various methods and techniques but they all failed. It might also be wrong ...
1
vote
1
answer
188
views
Can one show $h(x)=|2(\zeta'(x))^2-\zeta''(x)\zeta(x)|$ is a decreasing function for $x\in\mathbb{R}\cap [1,\infty)$?
This question is related to This question.
When I tried to approach it I couldn't even proof that the LHS is a decreasing function on the given domain using regular methods. I have tried to write the ...
2
votes
1
answer
160
views
Is the function $(z-1)(2^{z}-1)\zeta(z)$ logarithmically concave and convex in $z\in(0,\infty)$?
For proving that the sequence
\begin{equation}\label{first-proof-decreas-seq}
\frac{1}{(2k-1)(k+1)} \frac{2^{2k+2}-1}{2^{2k}-1} \biggl|\frac{B_{2k+2}}{B_{2k}}\biggr|
\end{equation}
is decreasing in $k\...
2
votes
0
answers
121
views
Solving a system of differential-like equations for reverse Euler-Maclaurin summation
Aim
A particular instance of a rational zeries that has as of yet not been evaluated is:
\begin{align}
Z:= \sum_{n=1}^{\infty} \frac{\zeta(2n)}{(2n)!}. \label{EM1} \tag{EM1}
\end{align}
This sum ...
1
vote
1
answer
101
views
Finding $\lim_{n \to \infty} \sum_{k=2}^{n-2} \zeta(k) \zeta(n-k) x^{k-1} = x^{-1} - \psi_{0}(-x) - \gamma$ from the generating function of $\zeta(•)$
In equation (130) of this page, the identity $$\lim_{n \to \infty} \sum_{k=2}^{n-2} \zeta(k) \zeta(n-k) x^{k-1} = x^{-1} - \psi_{0}(-x) - \gamma \label{1} \tag{1} $$ is stated. Here, $\zeta(\cdot)$ is ...
1
vote
0
answers
138
views
Integral of $|1/\zeta(\sigma+i T)|$ (or $|(1/\zeta(\sigma+i T))^{(k)}|$) on a horizontal half-line in the left upper quadrant
Let $T_0\geq 20$. Let $L$ be the half-line from $-\infty + i T$ to (say) $-1/2 + i T$. Since $|\zeta(s)|$ is roughly proportional to $(T/2 \pi e)^\sigma$ for $s=\sigma+ i T$ on $L$, it is clear that ...
3
votes
2
answers
597
views
lower bound for $\Re\zeta(1+it)$
Hi
is there any lower bound for $\Re\zeta(1+it)$.
I did try with computer until some ordinate and I saw $\Re\zeta(1+it)>0$.
If it is true, is there any reference to prove it.
thanks