@bobeyt6 From the description: "For feedback/discussion/requests of Close/Undelete/Reopen/Edit/Delete for questions and answers on Math SE (for Math Meta stuff go to chat.stackexchange.com/rooms/64952).";
Let $\sigma$ denote the divisor sum function.
Let $\theta$ be the supremum over all real parts of the non-trivial zeros of the zeta function, so $\theta>1/2$.
In Broughan's Equivalents of the Riemann Hypothesis
Theorem 7.11 on p.167 basically says the following:
Assume RH, then
$
\sigma(n) < e^\g...
Consider sequences $a_n$ for $n \geq 1$ taking values in $\pm1$. Call such a sequence multiplicative if $a_{mn}=a_ma_n$. I conjecture that if there exists such a multiplicative sequence $a_n$ and a real number $s \geq 0$ such that $\sum_{k=1}^{\infty}\frac{a_k}{k^s}=0$, (conditionally of course...
Thanks, @Peter, for the final delete on my first request here!
@XanderHenderson The post linked by Xander won't clear by Roomba, given an upvoted answer. I got it down to no votes; not sure whether letting it sit will be wise, given it was just closed yesterday. I'll keep my eye on it; in the meantime, more downvotes could help expedite its deletion. For reference: math.stackexchange.com./q/3255915
There's no quality control, when users over 300k continue to answer whatever they choose to answer, and are adorned with upvotes, even when answering questionable questions, and flagging re: EoQS of such users proves futile.
Consider sequences $a_n$ for $n \geq 1$ taking values in $\pm1$. Call such a sequence multiplicative if $a_{mn}=a_ma_n$. I conjecture that if there exists such a multiplicative sequence $a_n$ and a real number $s \geq 0$ such that $\sum_{k=1}^{\infty}\frac{a_k}{k^s}=0$, (conditionally of course...
