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+ prime numbers tag and minor formatting
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Myshkin
Myshkin
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implication Implications of divergence of $1/\zeta(s) $ at 1/2

$1/\zeta(s)=\sum_{n>0}\frac{\mu(n)}{n^s}$ where $\mu$ is the Moebius function. This

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).Convergence Convergence at 1 is equivalent to prime number theorem.Does

Does divergence at 1/2 have similiar implication.implications?

implication of divergence of $1/\zeta(s) $ at 1/2

$1/\zeta(s)=\sum_{n>0}\frac{\mu(n)}{n^s}$ 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).Convergence at 1 is equivalent to prime number theorem.Does divergence at 1/2 have similiar implication.

Implications of divergence of $1/\zeta(s) $ at 1/2

$1/\zeta(s)=\sum_{n>0}\frac{\mu(n)}{n^s}$ 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). Convergence at 1 is equivalent to prime number theorem.

Does divergence at 1/2 have similiar implications?

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Koushik
Koushik
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implication of divergence of $1/\zeta(s) $ at 1/2

$1/\zeta(s)=\sum_{n>0}\frac{\mu(n)}{n^s}$ 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).Convergence at 1 is equivalent to prime number theorem.Does divergence at 1/2 have similiar implication.