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Richard Borcherds
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$\sum_{n>0}\frac{\mu(n)}{n^s}$$1/\zeta(s)=\sum_{n>0}\frac{\mu(n)}{n^s}$ where $\mu$ is the Moebius function. It This series is known to converge for $s\ge 1$ and diverge for $s\le 1/2$. Its convergence is unknown if $1/2<s<1$, and convergence for all$1/2< s< 1$ $s>1/2$(convergence in this interval is equivalent toessentially the Riemann hypothesis).

$\sum_{n>0}\frac{\mu(n)}{n^s}$ where $\mu$ is the Moebius function. It is known to converge for $s\ge 1$ and diverge for $s\le 1/2$. Its convergence is unknown if $1/2<s<1$, and convergence for all $s>1/2$ is equivalent to the Riemann hypothesis.

$1/\zeta(s)=\sum_{n>0}\frac{\mu(n)}{n^s}$ where $\mu$ is the Moebius function. This series is known to converge for $s\ge 1$ and diverge for $s\le 1/2$. Its convergence is unknown if $1/2< s< 1$ (convergence in this interval is essentially the Riemann hypothesis).

Source Link
Richard Borcherds
  • 20.7k
  • 9
  • 97
  • 113

$\sum_{n>0}\frac{\mu(n)}{n^s}$ where $\mu$ is the Moebius function. It is known to converge for $s\ge 1$ and diverge for $s\le 1/2$. Its convergence is unknown if $1/2<s<1$, and convergence for all $s>1/2$ is equivalent to the Riemann hypothesis.