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seen Apr 15 at 12:57

Apr
13
revised The zeros of $2\,\xi(s)-1$. Is there anything known about the curves they lie on?
Fixed error. Forgot the factor s(s-1) on the right hand side in the second equation.
Apr
13
asked The zeros of $2\,\xi(s)-1$. Is there anything known about the curves they lie on?
Mar
21
comment Equality between an infinite product and an infinite series. How can I reconcile both?
@Gerry. I see that and believe your statement is indeed true for many cases, however there still is the option that two formulas evaluating to the same number actually do have a connection (with the Euler product as a trivial example).
Mar
21
revised Equality between an infinite product and an infinite series. How can I reconcile both?
Added a more worked out 'Euler product' example to explain the question further.
Mar
12
comment MoebiusMu product
@Fred. You're not talking to yourself but to myself now, right ? :-) Poor joke, I know. However, could you elaborate a bit on how you calculated the numerical convergence? Keen to reproduce it. Note that $5040 = 7!$.
Mar
5
comment Equality between an infinite product and an infinite series. How can I reconcile both?
@Lucian. Thanks. Studied both links, however still can't see a direct connection for the product/series with power 2. Note that the infinite products are 'alternating'. There is however a clear connection with the Wallis product for the power 1 (that I now added to my question).
Mar
5
revised Equality between an infinite product and an infinite series. How can I reconcile both?
Added a similar equation for the power 1 and an additional thought.
Mar
5
asked Equality between an infinite product and an infinite series. How can I reconcile both?
Mar
2
asked Question about the zeros and poles of the PrimeZeta function.
Feb
23
comment Convergence of a modified sum of prime reciprocals for all $s \in \mathbb{C}$?
Many thanks Daniel. Very helpful! Your logic obviously also applies to using integers $n$ instead of primes $p$ i.e. $\displaystyle f(s) := \sum^\infty_{n=2} \left( \frac{1}{n-\frac{1}{n^s}}- \frac{1}{n+\frac{1}{n^s}} \right)$ should converge in the same way. I experimented with $f(s)$ and found closed forms at integer values of $s$: $f(0)=-f(2)=\frac32$, $f(-1)=-f(3)=-\frac34 - i +\gamma + \Psi(i-1)-\frac{i}{2} \pi \coth(\pi)$, $f(-2)=-f(4)=\frac16$. After that the closed forms get complicated. Have not found similar closed forms for the primes.
Feb
23
accepted Convergence of a modified sum of prime reciprocals for all $s \in \mathbb{C}$?
Feb
23
asked Convergence of a modified sum of prime reciprocals for all $s \in \mathbb{C}$?
Jan
19
revised Are the zeros of the sum/difference of two reflexive, entire functions all on the line $\Re(s)=\frac12$?
Found counter examples so withdraw my claim.
Jan
19
asked Are the zeros of the sum/difference of two reflexive, entire functions all on the line $\Re(s)=\frac12$?
Jan
9
revised Infinite series with only two zeros at $\Re(s)=\frac12$. Why is that the case?
Added the comment about primes.
Jan
3
asked Infinite series with only two zeros at $\Re(s)=\frac12$. Why is that the case?
Dec
30
comment A function whose real part only converges (to zero) for $\Re(s)=\frac12$?
Many thanks, Giancane. That indeed seems to do the trick. This now shows that the summand of the infinite series in the linked question only converges for $\Re(s)=\frac12$, although I immediately have to make the caveat that the "limit of the summand"-test is inconclusive when the summand itself goes to zero... On the other hand, the latter does allow me to apply Leibniz's alternating series test, which implies that $\sum_{n=1}^{\infty}(-1)^{n} f(s,n)$ (i.e. ignore the limit for $n$ in $f(s)$) converges for $\Re(s)=\frac12$ only.
Dec
30
accepted A function whose real part only converges (to zero) for $\Re(s)=\frac12$?
Dec
30
revised A function whose real part only converges (to zero) for $\Re(s)=\frac12$?
Added brackets to avoid misunderstanding.
Dec
30
asked A function whose real part only converges (to zero) for $\Re(s)=\frac12$?